Prove that $A$ is closed I am asked to prove the following question:

Let "If $a \in \mathbb{R}$, $a_n \in A$ for all $n \in \mathbb{N}^+$, and $a_n \to a$, then $a \in A$." be a statement $P$. Show that if $P$ is true then $A$ is closed. 

I was given a hint that if $a \notin \text{int}(\mathbb{R} \setminus A)$, then for all $n \in \mathbb{N}^+$, there exists $a_n \in (a - 1/n, a+1/n) \cap A$. 

But I can't get my head around what is the hint trying to tell me and I don't know how to proceed. Can anyone give a hint? Thanks a lot!
I finally understood the question so here is an attempt at the proof.

Suppose $a \in \textbf{cl}A$, we have that $a \notin \textbf{int}(\mathbb{R} \setminus A)$. Thus, $a$ is not an interior point of $(\mathbb{R}\setminus A)$ and then we have for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \nsubseteq \mathbb{R}\setminus A$. This is equivalent to for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \cap A \neq \emptyset$. Hence, we let $a_n \in (a - 1/n, a + 1/n) \cap A$. Thus, $a_n \in A$ for all $n \in \mathbb{N}^+$. Moreoever, $a_n \to a$ since $\cap_{n = 1}^\infty(a- 1/n, a+ 1/n) = \{a\}$. Therefore, $a \in A$ by $P$. Hence, $\textbf{cl}A \subset A$ and $\textbf{cl}A = A$ since $A \subset \textbf{cl}A$. Therefore, $A$ is closed since $\textbf{cl}A$ is closed.
Hint: $a\not\in\operatorname{int}({\Bbb R\setminus A})$ is another way of saying $a\in\operatorname{cl}(A)$.  So, the hint is that every element of the closure of $A$ has a sequence in $A$ converging to it.  Now use $P$.