Proof involving a vacuously true statement Let $S$ be a finite subset of a metric space. Show that it is closed.
I know a set is closed if and only if it contains all of its accumulation points. Let $x$ be an accumulation point of $S$. I want to show that $x \in S$. Since $S$ is a finite set, I know that it cannot have a accumulation point. So does this imply that $x \in S$ and I am done with the proof?
If I have made an error, could someone explain? 

To prove that $S$ is closed in a metric space $M$, it is enough to show that $S$ contains all its accumulation points.
Since $S$ is finite, we can show that it has no accumulation points. Proof by contradiction: suppose that $S$ has an accumulation point $x$. Consider $r = \min |x-y|$ for all $y \in S$. Then the ball around $x$ with radius $r/2$ will contain no points of $S$. So $S$ contains no accumulation points. 
Since $S$ contains all of its accumulation points (since $S$ has none), that means $S$ is closed in $M$.
