Let $S$ be a connected metric space, with metric $ d:S\times S\to\mathbb R_{+} $. Let $ q\in S $. Show that if $S\backslash\{q\} $ is non-empty, then it is not compact.
Proof: Suppose the converse, i.e., $ S\backslash\{q\} $ is compact, then since a compact set in metric space is also closed, we deduce that $ S\backslash\{q\} $ is closed. Furthermore, since in $ T_1 $ space finite sets are closed, we get that $ \{q\} $ is closed. Hence $ S $ is the union of two disjoint non-empty closed sets $ S\backslash\{q\} $ and $ \{ q \} $, contradicting to $ S $ being connected.
Is it OK to give a proof like that?
Yes, the proof is fine; note also that we only need $S$ to be Hausdorff and crowded (no isolated points), both the metric and connectedness are overkill. If $S\setminus \{q\}$ is compact, your argument shows it's closed (in a Hausdorff space all compact subsets are closed) and so $\{q\}$ is open, i.e. $q$ is an isolated point, which a crowded space does not have. (a connected $T_1$ space is always crowded, but the reverse need not hold, e.g. consider the rationals as a subspace of the reals, which is totally disconnected, yet has no isolated points.)