To demonstrate that $ {x } $ is closed, one can show that $ X setminus {x } $ It's open.

Leave $ y in X setminus {x } $ and establish $ varepsilon = d (x, y) $ where $ d $ It is your given metric. But $ x not in B (and, varepsilon) $, Which means that $ B (y, epsilon) subset X setminus {x } $. But that clearly gives $ X setminus {a } $ is open, which implies that $ {x } $ The test is closed and completed.

**Note:** As Yanko pointed out in the comments, his given definition is incorrect.