By definition, $\mathbb{E}(X \mid \mathcal{G})$ is measurable in $\mathcal{G}$. Since $t \mapsto^2$ is a measurable transformation, then $(\mathbb{E}(X \mid \mathcal{G}))^2$ is also measurable in $\mathcal{G}$. Thus, applying results
we have
\begin{equation}
\begin{split}
\mathsf{Var}(X \mid \mathcal{G})
& = \mathbb{E}((X - \mathbb{E}(X \mid \mathcal{G}))^2 \mid \mathcal{G}) \\
& = \mathbb{E}(X^2 - 2 X \mathbb{E}(X \mid \mathcal{G}) + (\mathbb{E}(X \mid \mathcal{G}))^2 \mid \mathcal{G}) \\
& = \mathbb{E}(X^2 \mid \mathcal{G}) - 2 (\mathbb{E}(X \mid \mathcal{G})^2 + \mathbb{E}((\mathbb{E}(X \mid \mathcal{G}))^2 \mid \mathcal{G}) \\
& = \mathbb{E}(X^2 \mid \mathcal{G}) - 2 (\mathbb{E}(X \mid \mathcal{G})^2 + (\mathbb{E}(X \mid \mathcal{G}))^2 \\
& = \mathbb{E}(X^2 \mid \mathcal{G}) - (\mathbb{E}(X \mid \mathcal{G}))^2 \\
\end{split}
\end{equation}
$\square$