I know there is more than one question about this topic in the forum, but I would like to read a second opinion on my resolution of this exercise.
Problem. Given two (linear) compact operators $S,T\colon X \to Y,$ show that $S+T$ and $\alpha T$ are compact operators (for every $\alpha \in \mathbb K$).
My attempt. Firstly, consider the operator $S+T\colon X\to Y$ defined by $(S+T)x = Sx + Tx, \, \forall x \in X.$ Now, consider an arbitrary bounded sequence $(x_n)_{n \in \mathbb N} \subset X.$ Since $S$ is compact, the sequence $(Sx_n)_{n \in \mathbb N}$ admits a convergent subsequence (in $Y$) - let us call such subsequence by $(Sx_{n_i}),$ for some set of indexes $n_i \subset \mathbb N.$ Since the original sequence $(x_n)_{n \in \mathbb N}$ is bounded, it follows directly that the subsequence $(x_{n_i})$ is also bounded. Thus, since $T$ is a compact operator, the sequence $(Tx_{n_i})$ admits a convergent subsequence - let us call such subsequence by $T(x_{n_{i_j}}),$ for some set of indexes $n_{i_j} \subset n_i.$ Furthermore, we can also guarantee that the subsequence $(Sx_{n_{i_j}})$ is convergent just because $(Sx_{n_i})$ converges and $n_{i_j} \subset n_i.$ Since each of the subsequences $(Sx_{n_{i_j}})$ and $(Tx_{n_{i_j}})$ converges, it follows that the subsequence
$$ ((S+T)x_{n_{i_j}})$$ is also convergent (in fact, it converges to the sum of the limits of $(Sx_{n_{i_j}})$ and $(Tx_{n_{i_j}})$). But we just proved that $( (S+T)x_n)_{n \in \mathbb N}$ admits a convergent subsequence (in $Y$). Thus, $S+T$ is compact.
Now let us analyze the operator $\alpha T\colon X \to Y$ defined by $(\alpha T)x = \alpha Tx.$ Again, consider an arbitrary bounded sequence $(x_n)_{n \in \mathbb N} \subset X$. Since $T$ is compact, the sequence $(Tx_n)_{n \in \mathbb N}$ admits a convergent subsequence - let us call such sequence by $(Tx_{n_i}),$ for some set of indexes $n_i \subset \mathbb N.$ With this in mind, we can guarantee that the subsequence ´ $$ (\alpha Tx_{n_i}) = ((\alpha T)x_{n_i})$$ is also convergent (it converges to $\alpha$ times the limit of $(Tx_{n_i}).$ With this, we just proved that $((\alpha T)x_n)_{n \in \mathbb N}$ admits a convergent subsequence (in $Y$), proving that $\alpha T$ is compact.