An exercise that I am trying to tackle goes like this
$\ell^\infty$ defined as usual and a subset $S= \{ (x_n):|x_n| \leq \frac1n \text{ for all n}\}$
a) Prove that S is closed (Done, fairly straighforward)
b)Prove that S is totally bounded
(Hint: $\forall \epsilon>0, \text{consider the finite set } B_{\epsilon}=(\frac{\epsilon}{2}\Bbb Z)\cap [-1,1] \text{ and prove that } $
$S_{\epsilon}=\{(y_n): y_n\in B_{\epsilon} \text{ and } y_n=0 \text{ } \forall n>N_\epsilon\}\cap S $ is an $\epsilon$ -net for $S$ if $N_\epsilon$ is sufficiently large)
c) Prove that S is compact (can use that $\ell^\infty$ complete)
So part c) will be given after we have a) and b) since totally bounded and closed will imply that its compact (with some intermediate steps). I am not sure how to use the $\epsilon$-net hint that we are given.
$$\text{A subset of X and let }\epsilon>0 \text{ a finite set }A_\epsilon\text{ is an } \epsilon \text{-net} \text{ for A if A} \subset \cup_{x\in A_\epsilon}B^o(x,\epsilon)$$
The idea I think is pretty simple. Take $\epsilon > 0$. Take $N = N(\epsilon)$ so that $\frac{1}{N} < \epsilon$. Then if we just look at sequences which are $0$ from the $N^{th}$ coordinate on, the $L^\infty$ norm will be less than $\epsilon$ from any sequence in $S$, if we just consider the terms from the $N^{th}$ coordinate on. So we just need to worry about the first $N-1$ coordinates. And we can cover take care of these (with an $\epsilon$ of error) with only finitely many sequences. For the first coordinate, divide $[0,1]$ into $N$ equal pieces $I_1^{(1)},\dots,I_N^{(1)}$. Then for the second coordinate, divide $[0,\frac{1}{2}]$ into $N$ equal pieces $I_1^{(2)},\dots,I_N^{(2)}$. Etc until the $(N-1)^{st}$ coordinate. Then your $\epsilon$-net will be $(x_1,\dots,x_N,0,0,0,0\dots)$ where $x_1,\dots,x_N$ range over all the centers of the intervals $I_{j_1}^{(1)},I_{j_2}^{(2)},\dots,I_{j_N}^{(N)}$, respectively.