@Danielfischer For the proof of the first implication is the following fine:
Assume that there is a sequence in $A$ converging to $x$ but $x \notin cl(A)$. Hence $x$ is neither in $A$ nor a limit point of $A$. Since $x$ is not a limit point, there exists some neighbourhood $U$ of $x$ such that $U \cap A \setminus \{ x\} = \emptyset$, but this contradicts the assumption that $x_{n} \rightarrow x$, where $\{x_{n}\} \subset A$ unless $x_{n} = x$ for all $n \in \mathbb{N}$, but we also have that $x \notin A$, hence there is a contradiction. Therefore $x \in cl(A)$.