Is every finite subset of a metric space closed

Question

Is every finite subset of a metric space closed ? Yes/No

My attempt : i think No

Consider $X = \{0, 1\}$ with the indiscrete metrics i.e. the only two open sets are $\emptyset$ and $X = \{0, 1\}$ itself.

the singleton subset $\{0\}$ of $X$ is not closed because its complement $X \setminus \{0\} = \{1\}$ is not one of the two open sets listed above and therefore not open.

Is its true ?

Answer 1

Suppose $\{p_1,...,p_n\} \subset X$ and $x \notin F$. Let $\epsilon = \min_kd(x,p_k)$. Then $B(x,\epsilon) \cap F = \emptyset$ and so $F^c$ is open.