Suppose we have a metric space ($\Bbb{N}$,d) with metric $d(m,n)=\left|\frac{1}{n} - \frac{1}{m}\right|$
I need to prove that in this metric space, a set is sequentially compact $iff$ it is finite.
I think I have an idea of the the $\Rightarrow$ part of this iff:
Assuming that a set in this metric space is not finite and showing that we cant cover S with finitely many open open balls and that the next point in a set, say, {${x_1,x_2,...x_n}$} will have to be outside $\bigcup\limits_{i=1}^{n} B^o(x_i,\epsilon)$ which is a finite subcover. $d(x_n,x_m) \geq \epsilon$ and therefore { $x_n$} will not have a convergent subsequence and so we wont have a sequentially compact set. Contradiction.
For the $\Leftarrow$ part, I am not sure how I should start.
I tried saying that if S is finite then we have at least one of the values in S appearing infinitely many times so that we could have some sequence that has a convergent subsequence. But I am not sure how I can formalise on this.
Note: I can not use the fact that a set is compact iff it is sequentially compact.
Finite $\implies $ sequentially compact trivially, for every sequence would have a constant, hence convergent, subsequence...
This is the point you were trying to make...
This would be true in any topological space...