Why is the image of a compact operator separable?

• A countable union of separable sets is separable (if $S_j$ is separable, let $\left(x^{j}_n\right)_{n\geqslant 1}$ be a dense sequence in $S_j$; then $\left\{x_n^{j},n,j\geqslant 1\right\}$ is countable and dense in $\bigcup_{j\geqslant 1}S_j$).
• A subset $K$ with compact closure is separable (consider the covers $\left(B\left(x,n^{-1}\right)\right)_{x\in K}$).
• $A=\bigcup_{j\geqslant 0}B\left(0,j\right)$ and $S(A)=\bigcup_{j\geqslant 0}S\left(B\left(0,j\right)\right)$.

The space is a countable union of balls centered in zero: $$A = \bigcup_{n\in N}B(0,n).$$

The image of $B(0,n)$ is precompact, therefore, separable. Countable union of separable sets is separable.