I'm trying to proof that, given a metric space $(M, d)$ and $A \subseteq M$, the interior of A is a subset of the set of accumulation points of A. However I have apparently come across with a counterexample, although I think that it is very likely that I have made a mistake in some part.
Let $(\mathbb{N}, d_2)$ be a metric space and $A = \{5\} \subset \mathbb{N} $. $5$ is an interior point because if we choose $r = 1/2$ then $B_r(5) \subset A$. On the other hand, $B_r(5) \setminus \{5\} \cap \ A = \emptyset$, which means that 5 is an isolated point and, therefore, is not an accumulation point.
Where is the mistake?
Apparently, this was meant to be proved for $(M, d) = (\mathbb{R}^n, d_2)$.