Let $Y$ be a square integrable $(\mathcal F_t)_{t \in \mathbb R^+}$ martingale (namely that $E[Y^2_t]<\infty$ for all t) closed by $Y{_\infty}$.
$1)$ Then why do I know that also $E[Y_\infty^2]< \infty$?? Is it a consequence to the square integrability only or also of the fact that is closed by $Y{_\infty}$?
$2)$ Let $X_t=E[X_\infty|\mathcal F_t]$, then if I know that $X_\infty$is in $L^2$, why $X_t$ is a square integrable martingale?
If you just assume that $EY_t^{2} <\infty$ for each $t$ then $Y_t$ need not converge, so you don't have your $Y_{\infty}$. The first part can be approved under the assumption that $\sup_t EY_t^{2} <\infty$. When you say $(Y_t)$ is closed by $Y_{\infty}$ you probably mean $Y_t=E(Y_{\infty}|\mathcal F_t)$ for all $t$ with $Y_{\infty}$ square integrable. In that case we automatically get $\sup_t EY_t^{2} <\infty$.
Apply Fatou's Lemma: $EX_{\infty}^{2} \leq \lim \inf EX_t^{2}\leq \sup EX_t^{2} <\infty$.
The second part follows from Jensen's inequality for conditional expectations: $EX_t^{2}=E(E(X_{\infty}|\mathcal F_t)^{2}) \leq EX_{\infty}^{2}$ since $x \to x^{2}$ is a convex function.