Which of the following metric spaces $X$ are compact?
(a) Let $S$ is an infinite set, let $X$ be $\{ f:S \rightarrow \mathbb{R} \mid \| f\|_\infty \leq 1 \}$ equipped with sup norm
$$\|f-g\|_\infty = \sup_{s\in S}|f(s)-g(s)|.$$
(b) $X = \{(x_1,x_2,...) \mid x_n \in [0,\frac{1}{n}] \}$ with $l^2$ norm
$$d(x,y)=\sqrt{\sum(x_i - y_i)^2}.$$
To clarify what I have tried for (b), here is my attempt. I believe that there is a slicker method.
Proof. Let $((x_n)^{(k)})_{k\geq1}$ be a sequence in $X$. Consider $((x_1)^{(k)})_{k\geq1}$, since $[0,1]$ is compact, there exists a convergent subsequence $((x_1)^{(k_1)})_{k_1\geq1}$ which converges to $x_1$. Now consider $((x_2)^{(k_1)})_{k_1\geq1}$, similarly there exists a convergent subsequence $((x_1)^{(k_2)})_{k_2\geq1}$ which converges to $x_2$. Iterate this process.
We claim that the following subsequence $((x_n)^{(k_l)})_{l\geq1}$
$$(x_n)^{(k_1=1)}, (x_n)^{(k_2=2)}, (x_n)^{(k_3=3)}, ...$$
of the original sequence $((x_n)^{(k)})_{k\geq1}$ converges to $(x_n)$. Let $\epsilon > 0$, let $N$ be large so that
$$\sum _{n=N}^{\infty}\frac{1}{n^2} < \epsilon.$$
For $M$ large enough, for all $1 \leq i \leq N$,
\begin{alignat}{1} m \geq M &\implies |x_i^{(l_m)}-x_i|^2 < \frac{\epsilon}{i^2}\\ &\implies \sum_{i=1}^N |x_i^{(l_m)}-x_i|^2 \ < \ \sum_{i=1}^N\frac{\epsilon}{i^2}\ \leq \ \epsilon \frac{\pi^2}{6}\\ &\implies \sum_{i=1}^{\infty} |x_i^{(l_m)}-x_i|^2\ \leq \epsilon (\frac{\pi^2}{6} + 1). \end{alignat}
Hints: If you can find an infinite subset $\{a_n\}$ in your metric space such that $d(a_n, a_m) = 1$ for $n \ne m$, then your space is not compact. For part (b) a famous theorem in measure/integration theory may be useful.
Edit: Your proof for part (b) looks fine to me. You can avoid the estimates as follows: by Tychonoff, or by your diagonal subsequence argument, there is a subsequence of your original sequence in $X$ that converges pointwise. But every element in $X$ is bounded in absolute value by the $\ell^2$ sequence $(1/n)_{n \ge 0}$. Hence by the dominated convergence theorem, pointwise convergence implies $\ell^2$ convergence. But your estimates are not difficult, and in my opinion it is better to do it directly when you can.