let $K(X)$ denote the space of compact linear operators on a separable Banach space $X$ to itself. It is a known result that $K(X)$ is a closed subspace of $L(X)$, the space of linear operators on $X$ to itself, endowed with the operator norm.
Is it also known whether $K(X)$ is separable?
Let $X=\ell^1(\mathbb{N}).$ For any $a\in \ell^\infty(\mathbb{N})$ the operator $$T_ax=\left (\sum_{n=1}^\infty a_nx_n\right )\delta_1$$ is compact as $T_a$ is one-dimensional. Moreover the mapping $$\ell^\infty(\mathbb{N}) \ni a\longmapsto T_a \in K(X)$$ is an isometry. Hence $K(X)$ is not separable.
When $X=\mathcal{H}$ is a separable Hilbert space the finite dimensional operators are dense in $K(\mathcal{H})$ The $n$-dimensional operators form a subspace isomorphic to $\mathcal{H}^{2n}.$ Thus the space $K(\mathcal{H})$ is separable.