I am trying to prove that $\{2\}$ is both open and closed in $\{1,2,3\}$, equipped with the Euclidean metric. Is the following correct?
Openness: $B_{0.5}(2)\subset\{2\}$. Hence, $\{2\}$ is open in $\{1,2,3\}$.
Closedness: $\{2\}$ trivially contains all its limit points, since it has no limit points. Hence, $\{2\}$ is closed in $\{1,2,3\}$.
Thank you.
It's correct, and the answer is OK so long as you have already proven that $\{2\}$ has no limit points. This is simple to prove, of course, but it still should be done.
Alternatively, you could also say that because $\{1\}$ is also open (by the same argument as $\{2\}$ and $\{3\}$ is also open, then the complement of $\{2\}$, which is $\{1,3\}=\{1\}\cup\{3\}$, is open, which means $\{2\}$ is closed.