real analysis – Subset of discrete metric space is sequentially compact iff it’s finite

Hint: don’t worry about Cauchy sequences for this problem. Sequential compactness says that any sequence has a convergent subsequence without reference to the Cauchy property. Thus:

(1) If $(X, d)$ is a discrete metric space and $X$ is infinite, you can choose an infinite sequence $x_1, x_2, ldots in X$ such that, for any $i, j$, $x_i neq x_j$ unless $i = j$. Now you can show that $x_1, x_2, ldots$ has no convergent subsequence, because the only way a sequence in a discrete space can converge is if it is eventually constant.

(2) if $(X, d)$ is a discrete metric space and $X$ is finite, then any sequence $x_1, x_2 ldots in X$, must visit some $x$ in $X$ infinitely many times. so $x_1, x_2 ldots$ has a constant subsequence whose elements are equal to $x$ and therefore converges to $x$.