Let $X,Y$ be Banach spaces. Let $B(X,Y)$ be the set of bounded linear operators and let $K(X,Y)$ be the set of compact linear operators.
I want to prove that $K(X,Y)$ is a vector subspace of $B(X,Y)$. Please can you check my proof?
It is clear that $0$ is a compact operator. Hence $K(X,Y)$ is not empty and contains $0$.
Let $\lambda \in \mathbb C$, $u \in K(X,Y)$. Multiplication by $\lambda$ is a linear homeomorphism. Therefore, if $B$ denotes the unit ball in $X$, $\lambda\overline{ u(B)}$ is compact because $\overline{u(B)}$ is.
Let $u,v \in K(X,Y)$ and let $S$ be a bounded subset of $X$. The goal is to show that $\overline{u(S) + v(S)}$ is compact. To this end, note that closed subsets of compact sets are compact and that $u(S) + u(S) \subseteq \overline{u(S)} + \overline{v(S)}$. Hence it is enough to show that $\overline{u(S)} + \overline{v(S)}$ is compact:
Note that $\overline{u(S)} \times \overline{v(S)}$ is compact and that $+: Y \times Y \to Y$ is continuous hence it maps compact sets to compact sets hence $\overline{u(S)} + \overline{v(S)}$ is compact.