The question needs to be formulated more precisely. First a comment.
Any unbounded operator
$$ A: D(A)\subset E\to F $$
$E,F$ Banach spaces, can also be viewed as a bounded operator
$$ D(A)\to F $$
where $D(A)$ is equipped with the graph-norm $\vert-\vert_G$
$$\vert x\vert_G = \vert x\vert_E+\vert Ax\vert_F. $$
The operator is closed iff $(D(A),\vert-\vert_G)$ is Banach.
The above example illustrates the important role played by the domain of the operator.
Recall what closeness entails: if $(x_n)$ is a sequence in $D(A)$ such that
$$ x_n\to x_\infty\;\;\mbox{in}\;\;\vert-\vert_E, $$
$$ Ax_n \to y_\infty\;\;\mbox{in}\;\;\vert-\vert_F, $$
then $x_\infty \in D(A)$ and $y_\infty=A x_\infty$. Thus, the choice of the norms $\vert-\vert_{E,F}$ is also relevant to your question.
You mention that a differential operator with constant coefficients is unbounded and closed with respect to the uniform norm, but you give no suggestion of what its domain ought to be. To see that this issue is more subtle take a look at the following simple example.
Consider the operator $\newcommand{\pa}{\partial}$ $\newcommand{\bR}{\mathbb{R}}$ $A=\pa^2_x-\pa^2_y-\pa_z^2$ whose domain is a subspace $D(A)\subset C^0(\bR^3)$. Here I think that $E=F=C^0(\bR^3)$ with the usual $\sup$-norm. Can you think of a choice of $D(A)$ that will make this a closed operator closed?
Note that if you think of $A$ as an operator $C^2(\bR^3)\to C^0(\bR^3)$ it is obviously bounded. However, if you choose $D(A)= C^2(\bR^3)\subset C^0(\bR^3)$, it is not clear to me that the resulting operator
$$ A: D(A)\subset C^0(\bR^3)\to C^0(\bR^3), $$
is closed.
The above concrete operator was not chosen at random. The operator $A$ is a hyperbolic operator, and for such operators strange things could happen.
Best Answer
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.