Here are two approaches that you could take to compute the variance of $X_t$. I am not making the conditioning explicit as it just complicates the notation but doesn't really add any additional insights.
Compute $\mathbb{E} \left[ X_t \right]$ and $\mathbb{E} \left[ X_t^2 \right]$. You can then you use that
\begin{equation}
\text{Var} \left( X_t \right) = \mathbb{E} \left[ X_t^2 \right] - \mathbb{E} \left[ X_t \right]^2.
\end{equation}
Compute the cumulant generating function
\begin{equation}
\psi_{X_t}(\omega) = \ln \left( \mathbb{E} \left[ e^{\mathrm{i} \omega X_t} \right] \right).
\end{equation}
Then use that the $n$-th cumulant is linked to the $n$-th derivative of the cumulant generating function at $\omega = 0$
\begin{equation}
c_n \left( X_t \right) = \frac{1}{\mathrm{i}^n} \frac{\partial^n}{\partial \omega^n} \psi_{X_t}(0).
\end{equation}
The first two cumulants are equal to the mean and variance, respectively.
In what follows, I am working out the first approach. In your case, the dynamics of $X$ are given by
\begin{equation}
\mathrm{d}X_t = \lambda \left( 1 - X_t \right) \mathrm{d}t + \sigma \sqrt{X}_t \mathrm{d}W_t.
\end{equation}
Note that this is a special case of a square-root process with mean reversion speed $\lambda$ and a long-term mean of one. The solution here is thus a special case of the more general solution given e.g. in the context of the Cox-Ingersoll-Ross interest rate model or the Heston stochastic volatility model.
First, we remove the geometric drift term $-\lambda X_t \mathrm{d}t$ by setting $Y_t = e^{\lambda t} X_t$. Its differential is given by
\begin{eqnarray}
\mathrm{d}Y_t & = & \lambda e^{\lambda t} X_t \mathrm{d}t + e^{\lambda t} \mathrm{d}X_t\\
& = & \lambda e^{\lambda t} \mathrm{d}t + \sigma e^{\lambda t} \sqrt{X_t} \mathrm{d}W_t\\
& = & \lambda e^{\lambda t} \mathrm{d}t + \sigma e^{\lambda t / 2} \sqrt{Y_t} \mathrm{d}W_t.
\end{eqnarray}
Integrating yields
\begin{eqnarray}
Y_t & = & Y_0 + \lambda \int_0^t e^{\lambda u} \mathrm{d}u + \sigma \int_0^t e^{\lambda u / 2} \sqrt{Y_u} \mathrm{d}W_u\\
& = & Y_0 + e^{\lambda t} - 1 + \sigma \int_0^t e^{\lambda u / 2} \sqrt{Y_u} \mathrm{d}W_u.
\end{eqnarray}
Since $\mathbb{E} \left[ X_t \right] = e^{-\lambda t} \mathbb{E} \left[ Y_t \right]$, we conclude that
\begin{equation}
\mathbb{E} \left[ X_t \right] = e^{-\lambda t} X_0 + \left( 1 - e^{-\lambda t} \right).
\end{equation}
Next, we compute the differential of $Y_t^2$ and get
\begin{eqnarray}
\mathrm{d} Y_t^2 & = & 2 Y_t \mathrm{d}Y_t + \mathrm{d} \langle Y \rangle_t\\
& = & 2 \lambda e^{\lambda t} Y_t \mathrm{d}t + 2 \sigma e^{\lambda t / 2} Y_t \sqrt{Y_t} \mathrm{d}W_t + \sigma^2 e^{\lambda t} Y_t \mathrm{d}t.
\end{eqnarray}
Integrating again gives
\begin{equation}
Y_t^2 = Y_0^2 + \left( 2 \lambda + \sigma^2 \right) \int_0^t e^{\lambda u} Y_u \mathrm{d}u + 2 \sigma \int_0^t e^{\lambda u / 2} Y_u \sqrt{Y_u} \mathrm{d}W_u.
\end{equation}
The expectated value is
\begin{eqnarray}
\mathbb{E} \left[ Y_t^2 \right] & = & Y_0^2 + \left( 2 \lambda + \sigma^2 \right) \int_0^t e^{\lambda u} \mathbb{E} \left[ Y_u \right] \mathrm{d}u\\
& = & Y_0^2 + \left( 2 \lambda + \sigma^2 \right) \int_0^t e^{\lambda u} \left( Y_0 + e^{\lambda u} - 1 \right) \mathrm{d}u\\
& = & Y_0^2 + \left( 2 \lambda + \sigma^2 \right) \left( \frac{1}{\lambda} \left( Y_0 - 1 \right) \left( e^{\lambda t} - 1 \right) + \frac{1}{2 \lambda} \left( e^{2 \lambda t} - 1 \right) \right)
\end{eqnarray}
Since $\mathbb{E} \left[ X_t^2 \right] = e^{-2 \lambda t} \mathbb{E} \left[ Y_t^2 \right]$, we get
\begin{eqnarray}
\mathbb{E} \left[ X_t^2 \right] & = & e^{-2 \lambda t} X_0^2 + \left( 2 \lambda + \sigma^2 \right) \left( \frac{1}{\lambda} \left( X_0 - 1 \right) \left( e^{-\lambda t} - e^{-2 \lambda t} \right) + \frac{1}{2 \lambda} \left( 1 - e^{-2 \lambda t} \right) \right)\\
& = & e^{-2 \lambda t} X_0^2 + 2 X_0 \left( e^{-\lambda t} - e^{-2 \lambda t} \right) + \frac{\sigma^2}{\lambda} X_0 \left( e^{-\lambda t} - e^{-2 \lambda t} \right)\\
& & + \left( 1 - e^{-\lambda t} \right)^2 + \frac{\sigma^2}{2 \lambda} \left( 1 - e^{-\lambda t} \right)^2.
\end{eqnarray}
Since
\begin{eqnarray}
\left( \mathbb{E} \left[ X_t \right] \right)^2 = e^{-2 \lambda t} X_0^2 + 2 X_0 \left( e^{-\lambda t} - e^{-2 \lambda t} \right) + \left( 1 - e^{-\lambda t} \right)^2,
\end{eqnarray}
it follows that the variance of $X_t$ is given by
\begin{equation}
\text{Var} \left( X_t \right) = \frac{\sigma^2}{\lambda} X_0 \left( e^{-\lambda t} - e^{-2 \lambda t} \right) + \frac{\sigma^2}{2 \lambda} \left( 1 - e^{-\lambda t} \right)^2.
\end{equation}