Linear combinations of compact operators are compact

The zero operator is compact by [A2].

For bounded $E\subseteq X$ the sets $\overline{S(E)}$ and $\overline{T(E)}$ are compact by [A1], so their product is compact in $Y\times Y$ by [A4] and its image under the continuous addition map is compact by [A4]; since $(S+T)(E)\subseteq \overline{S(E)}+\overline{T(E)}$, its closure is a closed subset of that compact image, hence compact, and $S+T$ is compact.

For bounded $E\subseteq X$ the set $\lambda \overline{S(E)}$ is the image of the compact set $\overline{S(E)}$ under the continuous map $y\mapsto \lambda y$, hence compact by [A4]; since $(\lambda S)(E)\subseteq \lambda \overline{S(E)}$, its closure is compact by [A4], so $\lambda S$ is compact.

For every natural $j\ge 1$ the power $K^j$ is compact: $K^1=K$ is compact, and if $K^j$ is compact then $K^{j+1}=K\circ K^j$ is compact by [A3].

By [step 1.1], [step 1.2] and [step 1.3] the compact operators $X\to Y$ contain the zero operator and are closed under addition and scalar multiplication, so they form a linear subspace of $\mathcal{B}(X,Y)$.

Let $K$ be compact and $a_1,\dots ,a_m$ scalars with $m\ge 1$. Each $K^k$ with $1\le k\le m$ is compact by [step 1.4]; by induction on $m$ using [step 2.1], a finite sum of scalar multiples of compact operators is compact, so ${\sum }_{k=1}^m{a}_k{K}^k$ is compact.