Let's $W_t$ and $B_t$ are tow independent Brownian motion, where :

$W_t$ ~ $N(0, t)$,  
$B_t$ ~ $N(0, t)$  
$Cov(W_t,B_t)=0$

We know that sum of two Gaussian random variable is also Gaussian. 

$$E(1/2(W_t+B_t)) = 1/2(E(W_t+B_t))=0$$
$$Var(1/2(W_t+B_t))=1/4(var(W_t+B_t))=1/4(var(W_t)+var(B_t))=.5t$$ 

because $W_t$ and $B_t$ are independent. So: 
 
$X_t$=$1/2(B_t+W_t)$ ~ $N(0, .5t)$

EDIT : $X_t$ has continuous path and $X_t=0$ for $t=0$ but $Var(X_t) \neq t$( a necessary condition for Brownian Motion). Hence $X_t$ is not Brownian Motion.

@Gordon mention rightly $\sqrt{1/2}(Wt+Bt)$ is a BM but not $X_t$.