'Suppose $X$ is a compact metric space and then $X$ is totally bounded"
Definition:
A metric space $X$ is said to be totally bounded if $\forall \epsilon>0$, there is a finite set $\{s_1,...,s_n\}$ such that $X=\cup_{i=1}^{n}B(s_i,\epsilon)$
Compact means sequential compactness rather than in terms of finite sub cover.
Proof:
Fix $\epsilon >0$ then suppose for a contradiction that $X$ is not totally bounded then we may find the set $Y=\{s_1,...,s_n\}$ such that $d(s_i,s_j)>\epsilon$. Since $X$ is not totally bounded, we can find $s_{n+1}$ such that $d(s_{n+1},s_j)>\epsilon$ for all j less than $n+1$. Doing this inductively, we see that we can not find a subsequence that converges.
My confusion:
Please see the line that is in italic, why can we always such $s_{n+1}$? it doesn't seem too trivial for me to understand, I assume it has something to do with the fact that by assumption $X$ is not totally bounded? (I would understand that there always exists s_{n+1} such that $d(s_{n+1},s_j)>\epsilon$ for some j)
If $X$ is not totally bounded then we can take some $e>0$ such that no finite set of open $e$-balls covers X.
Take $x_1\in X.$ Recursively for $n\in \Bbb N,$ since $X\ne \cup_{j=1}^nB(x_j,e),$ take $x_{n+1}\in X\setminus \cup_{j=1}^nB(x_j,e).$
Now if $m,n \in \Bbb N$ with $m<n$ then $x_n\not \in \cup_{j=1}^{n-1}B(x_j,e)\supset B(x_m,e),$ so we have $$(I)....\quad m\ne n\implies d(x_m,x_n)\ge e.$$ So $(x_n)_{n\in \Bbb N}$ has no convergent sub-sequence. For if $x$ were the limit of a convergent sub-sequence, then there would be infinitely many $n$ such that $d(x,x_n)<e/2.$ But if $d(x,x_{n_1})<e/2$ and $d(x,x_{n_2})<e/2$ with $n_1\ne n_2$ then $d(x_{n_1},x_{n_2})<e$ contrary to $(I).$
So if $X$ is not totally bounded then $X$ is not sequentially compact.