An operator $A: D(A)\to H$ of an Hilbert space $H$ is closed if his graph is closed (in the product topology).
An operator is closable if there exists an extension such that his graph is the closure of the graph of the operator.
I want prove that if $A: D(A)\to H$ is an operator for which it is true the property
for every $x_ny_n\subset D(A)$ such that $x_n,y_n\to x\in H$ and such that $Ax_n$ and $Ay_n$ are convergent then $\lim_{n}Ax_n=\lim_{n}Ay_n$
then A is closable.
My idea is to define
$D(B):=\{x\in H: \text{ exists } x_n\subset D(A) \text{ for which } $
$x_n\to x \text{ and } $Ax_n $ \text{ is convergent } \}$
and $B: D(B)\to H$ that maps every $x\in D(B)$ to $Bx=\lim_{n\to \infty}Ax_n$.
This map is well defined for the property that i requested.
Oviously the map is an extension of $A$ but i don’t know how to prove that his graph is closed.
The graph of your operator $B$ is precisely the closure of the graph of $A$. Indeed, we have \begin{align*} G(B) &= \{(x,y) \in H^2: \text{ there exists } x_n \in D(A) \text{ such that } x_n \to x, Ax_n \to y\} \\ &= \{(x,y) \in H^2: \text{ there exists } (x_n, Ax_n) \in G(A) \text{ such that } (x_n, Ax_n) \to (x,y) \} \\& = \overline{G(A)} \end{align*}
In general, we have that $A$ is closable if and only if $\overline{G(A)}$ is the graph of some operator. In turn, a useful general fact is that this happens if and only if whenever $x_n \in D(A)$ and $x_n \to 0$, $Ax_n \to y$ we have that $y=0$. If you know this general criterion, it's even easier to prove your result.
Let $x_n$ be a sequence in $D(A)$ such that $x_n \to 0$ and $Ax_n \to y$. Then $$y = \lim_{n \to \infty} Ax_n = \lim_{n \to \infty} A(0) = 0$$ by the given assumption. Hence, $A$ is closable.