Although Math SE might be a bit more suited for this one, I wanted to give it a try.
The answer relies on the law of total expectation, the law of total variance, and the relationship between Euler's number $e$ and series involving the factorial function $n!$.
In the first half of the answer, I will present the derivation of $E(n)$ and $Var(n)$, i.e. mean and variance of the stopping time, in the second half I will show how to derive $E(x_n)$ and $Var(x_n)$, i.e. the expected value and variance of the stopped process. Finally, in the addendum, I provide the full distribution of the stopped process, $f(x_n)$ for $x_k$ drawn from a uniform and from a generic distribution.
Part I: The distribution of the stopping time $n$
Given the $n$th draw $x_{n}$ from the unit interval $(0,1)$, we accept the ($n+1$)th draw $x_{n+1}$ if $x_n<x_{n+1}$; else we stop at the $n$th draw and the process is said to be stopped at $n$ with value $x_n$. We say that $n$ is the stopping time.
Let’s say we are about to draw $k$ numbers $x_1, x_2, \ldots x_k$ uniformly from $(0,1)$ and arrange the results in increasing order, e.g.
$$
\begin{align}
x_1&<x_2<\ldots<x_k
\end{align}
$$
Then there are $k!=k\times(k-1)\times(k-2)\ldots\times2\times1$ possible permutations of the ordering of the draws, each permutation with the same probability $1/k!$ - but only one permutation represents the necessary ordering $x_1<x_2<\ldots<x_k$ under which the process is not yet stopped at draw $k$ (of course, it will be stopped if $x_k>x_{k+1}$). Thus, the unconditional probability of $n\geq k$, i.e. of drawing at least $k$ times, is
$$
\begin{align}
P(n\geq k)&=\frac{1}{k!} \\
\Rightarrow P(n=k)&=P(n\geq k)-P(n\geq k+1)\\
&=\frac{1}{k!}-\frac{1}{(k+1)!}
\end{align}
$$
Calculating the expected stopping time
We calculate $E(n)=\sum_k k\times P(n=k)$, rearranging the factorials in the meantime:
$$
\begin{align}
E(n)&=\sum\limits_{k=1}^\infty \frac{k}{k!}-\frac{k}{(k+1)!}\\
&=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\ldots\right)-\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots\right)\\
&=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots\\
&=\sum\limits_{k=1}^{\infty}\frac{1}{k!}\\
&=\sum\limits_{k=0}^{\infty}\frac{1}{k!}-1
\end{align}
$$
In the last row, we recognize the definition of Euler's constant $e=\sum_k 1/k!=2.718282\ldots$; hence:
$$
E(n)=e-1
$$
Calculating the variance of the stopping time
As $Var(x)=E(x^2)-E(x)^2$, we now try to estimate $E(n^2)$, again applying some rearranging, and observing relationships with the definition of $e$:
$$
\begin{align}
E(n^2)&=\sum\limits_{k=1}^\infty k^2\times P(n=k)\\
&=\sum\limits_{k=1}^\infty\frac{k^2}{k!}-\frac{k^2}{(k+1)!}\\
&=\left(\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+\ldots\right)-\left(\frac{1^2}{2!}+\frac{2^2}{3!}+\frac{3^2}{4!}+\ldots\right)\\
&=\sum\limits_{k=1}^{\infty}\frac{k^2-(k-1)^2}{k!}\\
&=\sum\limits_{k=1}^{\infty}\frac{2k-1}{k!}\\
&=2\sum\limits_{k=1}^{\infty}\frac{k}{k!}-\left(e-1\right)\\
&=2\sum\limits_{k=0}^{\infty}\frac{1}{k!}-\left(e-1\right)\\
&=e+1
\end{align}
$$
Thus
$$
\begin{align}
Var(n)=&E(n^2)-E(n)^2\\
&=e+1-(e-1)^2\\
&=3e-e^2
\end{align}
$$
A simulation of the stopping time $n$
Let's simulate this using R. I choose a cutoff level n=14, the corresponding probability is below 1E-10.
set.seed(42)
nSim <- 1E6
k <- 14
mu <- exp(1)-1
v <- 3*exp(1)-exp(2)
u<-sapply(1:nSim,function(i){
x <- rle( diff( runif(k) ) > 0 )
1 + x$values[1] * x$lengths[1]
})
print( mean(u)- mu )
print( var(u) - v )
with output
[1] -0.0001078285
[1] -0.0003605153
Part II: Expectation and variance of $x_n$
Here, we will make use of the law of total expectation and the law of total variance.
Expectation of $x_n$
We start with the law of total expectation, applying it to $x_n$ as
$$
\begin{align}
E(x_n)&=E\left(E\left(x_n|n=k\right)\right)\\
&=\sum\limits_{k=1}^{\infty}E\left(x_n|n=k\right)P(n=k)
\end{align}
$$
We already know the distribution of $n$, all that remains is calculating $E(x_n|n=k)$.
For a (fixed) stopping time $k$, it must hold that $x_1\leq x_2\leq\ldots \leq x_{k-1}\leq x_k > x_{k+1}$, i.e. the $k$th component from $k+1$ draws represents the maximum over $(k+1)$ draws. The distribution of the maximum is:
$$
\begin{align}
F^{max}_{k+1}(x)&\equiv P(\max(x_1,x_2,\ldots,x_k,x_{k+1})\leq x)\\
&=P(x_1\leq x, x_2\leq x, \ldots x_{k-1}\leq x,x_{k+1}\leq x)\\
&=F(x)^{k+1}\\
&=x^{k+1}
\end{align}
$$
The corresponding density, $f^{max}_{k+1}(x)=\partial F^{max}_{k+1}(x)/\partial x$, is
$$
f^{max}_{k+1}(x)=(k+1)x^k
$$
and the expectation of $x_n$ given stopping at $k$ is
$$
\begin{align}
E(x_n|n=k)&=\int\limits_0^1 xf^{max}_{k+1}(x)\mathrm{d}x\\
&=(k+1)\int\limits_0^1 x^{k+1}\mathrm{d}x\\
&=\frac{k+1}{k+2}
\end{align}
$$
We now have both ingredients for the expectation:
$$
\begin{align}
E(x_n)&=\sum\limits_{k=1}^{\infty}E\left(x_n|n=k\right)P(n=k)\\
&=\sum\limits_{k=1}^{\infty}\frac{k+1}{k+2}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\
&=\left(\frac{2/3}{1!}+\frac{3/4}{2!}+\frac{4/5}{3!}+\frac{5/6}{4!}+\ldots\right)-\left(\frac{2/3}{2!}+\frac{3/4}{3!}+\frac{4/5}{4!}+\frac{5/6}{5!}+\ldots\right)\\
&=0.5+\sum\limits_{k=1}^{\infty}\frac{1}{(k+2)!}\\
&=0.5+\sum\limits_{k=1}^{\infty}\frac{1}{k!}-1.5\\
&=\sum\limits_{k=1}^{\infty}\frac{1}{(k)!}-1\\
&=e-2
\end{align}
$$
where we have reused the relationships between the inverse factorial sum and Euler's number, above.
Variance of $x_n$
We make use of the law of total variance
$$Var(x_n)=E(Var(x_n|n=k))+Var(E(x_n|n=k))$$
and start with the first term:
$$
\begin{align}
E(Var(x_n|n=k))&=\sum\limits_{k=1}^{\infty}\left(E(x_n^2|n=k)-E(x_n|n_k)^2\right)P(n=k)\\
&=\sum\limits_{k=1}^{\infty}\left((n+1)\int_{x=0}^1x^2f^{max}_{k+1}(x)\mathrm{d}x-\left(\frac{k+1}{k+2}\right)^2\right)P(n=k)\\
&=\sum\limits_{k=1}^{\infty}\left(\frac{k+1}{k+3}-\left(\frac{k+1}{k+2}\right)^2\right)P(n=k)
\end{align}
$$
and stop here. For the second term, we get
$$
\begin{align}
Var(E(x_n|n=k))&=E(E(x_n|n=k)^2)-E(E(x_n|n=k))^2\\
&=\sum_{k=1}^{\infty}\left(\frac{k+1}{k+2}\right)^2P(n=k)-\left(\sum_{k=1}^{\infty}\frac{k+1}{k+2}P(n=k)\right)^2\end{align}
$$
Combining both terms yields
$$
\begin{align}
Var(x_n)&=E(Var(x_n|n=k))+Var(E(x_n|n=k))\\
&=\sum\limits_{k=1}^{\infty}\frac{k+1}{k+3}P(n=k)-\left(\sum_{k=1}^{\infty}\frac{k+1}{k+2}P(n=k)\right)^2
\end{align}
$$
which, after some more rearranging yields
$$
Var(x_n)=2+2e-e^2
$$
A simulation study of $x_n$
set.seed(42)
nSim <- 1e6
k <- 14
u <- sapply(1:nSim,function(i){
x <- runif(k)
z <- rle(diff(x)>0)
x[1 + z$values[1]*z$lengths[1]]
})
cat("Variable : " ,"\t", "simulation", "\t", "theoretical" , "\n")
cat("mean(x_n) : " ,"\t", mean(u) , "\t", exp(1)-2 , "\n")
cat("var(x_n) : " ,"\t", var(u) , "\t", 2+2*exp(1)-exp(2), "\n")
resulting in
Variable : simulation theoretical
mean(x_n) : 0.7182006 0.7182818
var(x_n) : 0.04752103 0.04750756
Addendum
We can even produce the distribution of $x_n$ in closed form by combining the PMF of $n$ and the density $f^{max}_{k+1}(x)$:
$$
\begin{align}
f(x)&=\sum\limits_{k=1}^{\infty}f(x_n|n=k)P(n=k)\\
&=\sum\limits_{k=1}^{\infty}(k+1)(x)^k\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\
&=\sum\limits_{k=1}^{\infty}(x)^k\frac{1}{(k-1)!}\\
&=x\sum\limits_{k=0}^{\infty}(x)^k\frac{1}{k!}\\
&=xe^{x}
\end{align}
$$
By the same line of reasoning we can show that the distribution of $x_n$ given draws from a continuous distribution $\Phi$ (and corresponding density $\phi$) equals
$$
f(x)=\phi(x)\Phi(x)e^{\Phi(x)}
$$
