Let $X$ be a Banach space, $S=\left\{x_n\right\}_{n\in \mathbb{N}}\subset X$ a pre-compact and countable set. Is there an easy way, heavily based on the fact that $S$ is countable, to see that its convex hull, $co(S)$, is still pre-compact? That is, that every sequence $\left\{y_k\right\}_{k\in \mathbb{N}}\subset co(S)$ has a converging subsequence in $X$?
I know that more generally the convex hull of a pre-compact subset of a Banach space is pre-compact. However I was trying to find an alternative solution to this problem by reducing it to the case where the subset is also countable, thanks to the fact that any precompact subset of a Banach space is totally bounded and hence separable.
I tried to reduce the problem to the countable case because I initially thought it could then be solved easily using diagonalization. Expressing each $y_k$ as a convex combination $$y_k=\sum_{n=1}^{N_k}\lambda_{n,k}x_n,\qquad \lambda_{n,k}\in [0,1],\;\sum_{n=1}^{N_k}\lambda_{n,k}=1 $$ Here each $\left\{\lambda_{n,k}x_n\right\}_{k\in \mathbb{N}}$ has a converging subsequence. Then diagonalization allows me to prove that there is a subsequence $\left\{y_k\right\}_{k\in I}$, indexed by $I\subset \mathbb{N}$, such that $\left\{\lambda_{n,k}x_n\right\}_{k\in I}$ converges for all $n$. But then I still need to sum over $n=1,\dots, N_k$ and $N_k$ depends on $k$ so I cannot deduce convergence for $\left\{y_k\right\}_{k\in I}$. So, I am starting to think that diagonalization does not work here, but I am still a novice with this technique so I may not be applying it in the right way.
There are various posts on stackexchange giving proofs of the fact that the convex hull of a precompact set in a Banach space is precompact. The proof is not difficult, and the diagonalization argument you have in mind is very likely to lead to a longer proof.
Anyway, what you have in mind can be done when the sequence $\{x_n\}$ converges to some $x$. In the case $x=0$ this is done in Lemma 3.4.29 of Megginson's book "An introduction to Banach space theory". The proof is due to Grothendieck. The generalization to $x\ne 0$ now is trivial.