Let $\ell_2:=\{ x:\mathbb{N} \to \mathbb{R}: \sum_{j=1}^\infty x_j^2<\infty\}$ and $c_0:=\{ x:\mathbb{N} \to \mathbb{R}: \lim_{j\to\infty}x_j=0,\, \sup_{j\in\mathbb{N}}|x_j|<\infty\}$ denote the usual Banach sequence spaces. Given Banach spaces $A,B$, let $L(A,B)$ denote the Banach space of bounded, linear operators from $A$ to $B$.
Question: is $L(\ell_2,\ell_2)\subseteq L(\ell_2,c_0)$ dense?
Choose finite sets $A_n\subset{\bf N}$ with $\max(A_n)<\min(A_{n+1})$, and then define $T:c_{00}\to c_0$ by $T(e_n)=\chi_{A_n}$.
It is easy to check that $T$ extends to a norm 1 linear map $\ell^2\to c_0$. On the other hand, I think that if you take $|A_n|\to\infty$ fast enough, then for each norm 1 $R:\ell^2\to\ell^2$ you can find $\xi\in c_{00}$ such that $\xi$ has 2-norm equal to 1 but $R\xi-T\xi$ has $c_0$-norm at least 1/3. Now use Riesz's lemma to conclude that the answer to your question is negative.
(I am writing in a hurry on an iThing, so apologies for lack of detail and pretty formatting. When I have breathing space at a keyboard I will try to elaborate.)
Added later:
As Mikael has pointed out in comments, one does not need any assumption on the growth rate of $|A_n|$, merely that $\sup_n |A_n|=+\infty$. Moreover, one doesn't need Riesz's lemma.
To give a little more detail: suppose $R: c_{00}\to \ell^2$ is any linear map such that $\Vert R - T \Vert_{2\to \infty} = \varepsilon < 1$. Then for every $k\in A_n$ we have $|(R e_n)_k| \geq 1-\varepsilon$, and thus $$ \Vert R e_n \Vert_2 \geq (1-\varepsilon) |A_n|^{1/2} $$ Provided that $\sup_n |A_n|=+\infty$ this shows that $R$ cannot extend to any continuous linear map $\ell^2\to\ell^2$.
Remark you can all skip: The reason I originally thought we might want growth conditions on the sequence $(|A_n|)_{n\geq 1}$ is because I was misremembering something I'd worked out years ago. Namely, if $1\leq p < \infty$ and we let $B$ be the subalgebra of $L(c_0,c_0)$ consisting of all operators that moreover map the dense subspace $\ell^p\subset c_0$ to itself, then $B$ is not dense in $L(c_0,c_0)$. (This rediscovery is what prompted me to ask this old MO question.) Explicit operators in $L(c_0,c_0)$ which can't be approximated by elements of $B$ can be provided by spreading shifts of the kind used in my answer here, but the proof they are not in the closure of $B$ seems to require more work. However, I'm not sure if this question is genuinely related to the one you asked, or whether they are just different questions involving some of the same symbols.
