The set $Ext(X_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)

Suppose that $X$ is a Banach space which is complemented in $X^{**}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X_0$ say. Then there is a compact operator $K\colon X_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 here; apologies for self-advertisement). On the other hand, $X_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable. The spaces $\ell_p$ for $p\in [1,2)\cup (2,\infty)$ together with $L_1$ satisfy the above assumptions.

Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}_\infty$-space as in this case we have the Grothendieck–Lindenstrauss theorem about extensions of compact operators.

The set $Ext(X_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)

Suppose that $X$ is a Banach space which is complemented in $X^{**}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X_0$ say. Then there is a compact operator $K\colon X_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 here; apologies for self-advertisement). On the other hand, $X_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable. 

The classical spaces $\ell_p$ and $L_p$ for $p\in [1,2)\cup (2,\infty)$ satisfy the above assumptions. The assumption that $X\cong X\oplus X$ can be dropped but things get messier then.

Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}_\infty$-space as in this case we have the Grothendieck–Lindenstrauss theorem about extensions of compact operators.

The set $Ext(X_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)

Suppose that $X$ is a Banach space which is complemented in $X^{**}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X_0$ say. Then there is a compact operator $K\colon X_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 here; apologies for self-advertisement). On the other hand, $X_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable.

Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}_\infty$-space as in this case we have the Grothendieck–Lindenstrauss theorem about extensions of compact operators.

The set $Ext(X_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)

Suppose that $X$ is a Banach space which is complemented in $X^{**}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X_0$ say. Then there is a compact operator $K\colon X_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 here; apologies for self-advertisement). On the other hand, $X_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable. The spaces $\ell_p$ for $p\in [1,2)\cup (2,\infty)$ together with $L_1$ satisfy the above assumptions.

Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}_\infty$-space as in this case we have the Grothendieck–Lindenstrauss theorem about extensions of compact operators.