Prove that a metric space $(X,d)$ is sequentially compact iff every infinite subset has a cluster point.
Assume there is some infinite subset $S$ that has no cluster points in $X$, that is for every $s\in S$ there is some open ball $B(s,\epsilon_s)$ that contains only finitely many points from $X$. Since $S$ is infinite, we can choose some infinite sequence of points $(s_n)$ from it, but it can't contain any convergent subsequence, contradiction.
Assume that every sequence in $X$ has a convergent subsequence. Then also every sequence from an infinite subset $S$ of $X$ has a convergent subsequence. Thus it has a cluster point, i.e. limit of that subsequence.
Could someone check this "proof"?