Let $\left\{X_{n}, \mathcal{F}_{n}\right\}_{n \geq 0}-$ square integrable martingale, and $$ \sum_{n \geq 1} \mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}<\infty. $$
If exists Random variable $X_\infty$.
Want to prove $$ \mathrm{E} X_{\infty}^{2}<\infty, \quad X_{n} \rightarrow X_{\infty} \text { a. s. } ,\quad \mathrm{E}\left(X_{n}-X_{\infty}\right)^{2} \rightarrow 0 $$
Here is the full answer note that we need to use martingale property
So as mentioned in my comment above under hypothesis lets show that $X_n$ is Cauchy in $L^2$ which is a Banch space and os is complete : $$\sum_{n \geq 1} \mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}<\infty$$
So at $\epsilon >0$ fixed there exist $N_0$ s.t. for $n,p > N_0$:
$$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}+X_{n-1}- ... +X_{p+1}-X_{p}\right)^{2}=\sum_{i=p+1}^n \mathrm{E}\left(X_{i}-X_{i-1}\right)^{2}$$
the reason for that is the following : if $n=p+1$ then : $$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left((X_{n}-X_{n-1})+(X_{n-1}- X_{n-2})\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}+$$ $$\mathrm{E}\left(X_{n-1}-X_{n-2}\right)^{2}+ 2.\mathrm{E}\left(X_{n}-X_{n-1}\right).\left(X_{n-1}-X_{n-2}\right) $$ But $\mathrm{E}\left(X_{n}-X_{n-1}\right).\left(X_{n-1}-X_{n-2}\right)=0$ as from martingale property : $\mathrm{E}(X_{n}-X_{n-1}).(X_{n-1}-X_{n-2})=\mathrm{E}(X_{n}X_{n-1}) - \mathrm{E}(X_{n-1}^2)+\mathrm{E}(X_{n-1}X_{n-2})-\mathrm{E}(X_{n}X_{n-2})$
$$=\mathrm{E}(\mathrm{E}(X_{n}|\mathcal{F}_{n-1})X_{n-1}) - \mathrm{E}(X_{n-1}^2)$$ $$+\mathrm{E}(\mathrm{E}(X_{n-1}|\mathcal{F}_{n-2})X_{n-2}) -\mathrm{E}(\mathrm{E}(X_{n}|\mathcal{F}_{n-2})X_{n-2})$$ $=\mathrm{E}(X_{n-1}.X_{n-1}) - \mathrm{E}(X_{n-1}^2)$ (using (1) below)
$+\mathrm{E}(X_{n-2}.X_{n-2}) -\mathrm{E}(X_{n-2}.X_{n-2})$ (using (1) below) $$=0+0$$
So by recurrence and as the series is convergent we are assured that at fixed $\epsilon>0$ there exists some $N_0$ such that for every $n,p> N_0$ we have :
$$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}+X_{n-1}- ... +X_{p+1}-X_{p}\right)^{2}=\sum_{i=p+1}^n \mathrm{E}\left(X_{i}-X_{i-1}\right)^{2}<\epsilon $$
The $X_n$ is Cauchy in a complete space $L^2$, which entails existence of a limit $X_\infty$ up to almost sure equivalence class and everything in your conclusion holds true without further ado.
(*) This is true as for $n>k$ and $X \in \mathcal{F}_{n}$, the conditional expectation of $X$ knowing $\mathcal{F}_{k}$ is the (only) r.v. $Z\in\mathcal{F}_{k}$ s.t. for every $Y$ in $\mathcal{F}_{k}$ we have :
$$\mathrm{E}(Z.Y)=\mathrm{E}(X.Y)$$ (1)
Then such a $Z$ is denoted $\mathrm{E}(X|\mathcal{F}_{k})$ and it's unique almost surely.
Regards