I have the following proof in my notes:
T and S are compact operators. We need to proof that a linear combination is still a compact operator Let $\{x_n\}$ be a bounded sequence in $X$, then because $T$ is compact there exists a convergent subsequence $\{Tx_{n_k}\}$ . Because $S$ in compact, there exists a convergent subsequence $\{Sx_{n_{k_j}}\}$. Hence, $\alpha Tx_{n_{k_j}} + \beta Sx_{n_{k_j}}$ is convergent.
I'm lost after getting $\{Tx_{n_k}\}$ . In particular I don't understand why they are taking a subsequence of a subsequence here $\{Sx_{n_{k_j}}\}$, since I don't know if $\{Tx_{n_k}\}$ or $\{Sx_{n_k}\}$ are bounded in order to extract a subsequence of that subsequence. I only know that $\{x_n\}$ is bounded.
I would do it like this: I can say for sure is that there exist convergent subsequences $\{Tx_{n_k}\}$ and $\{Sx_{n_k}\}$. Then I would say that the linear combination $\alpha Tx_{n_k} + \beta Sx_{n_k}$ is convergent so I have a convergent subsequence of $\{(\alpha T + \beta S)x_n\}$ , so $\alpha T + \beta S$ is compact. Isn't this correct?
No, your argument is not correct. You can only say that $Tx_{n_k}$ is convergent for some $(n_k)$ and $Sx_{m_k}$ is convergent for some $(m_k)$. You do not know that you get the same subsequence for $T$ and $S$.
After getting $x_{n_k}$ you look at the new bounded sequence $(x_{n_k})$ and use compactness of $S$ to get a further subsequence $(n_{k_j})$ with $(Sx_{n_{k_j}})$ convergent. Then note that $(Tx_{n_{k_j}})$ is subsequence of $(Tx_{n_k})$ so it is also convergent. Now you see that $\alpha Tx_{n_{k_j}}+\beta Sx_{n_{k_j}}$ is convergent.