Interior is a Subset of the Closure 
$\textbf{Question:}$ Let $A$ be a set. Why is the $interior(A)\subset closure(A)$?

I use the following definitions for $\underline{interior}$ and $\underline{closure}$ given a metric space $X$.


*

*Interior Definition: $x\in interior(A)$ if $\exists$ a neighborhood
$N$ of the point $x$ such that $N\subset A$.

*Closure Definition: $x\in closure(A)$ if $\forall$ neighborhoods $N$
of the point $x$, $N\cap A\neq \phi$.


I started trying to prove this as follows, any help would be greatly appreciated!
Proof: Let $x\in interior(A)$. Thus, $\exists N$ of $x$ such that $N\subset A$. Say $N_0$ is such a neighborhood of $x$ which $N_0\subset A$ holds true. Then, I get stuck here as I am trying to show $x\in closure(A)$ by showing $\forall N$ of $x$, $N\not\subset A$ holds true or $N\cap A\neq \phi$ holds true (these are equivalent).

Hint: Rather prove $\mathrm{int}(A) \subseteq A\subseteq\mathrm{clo}(A)$. 

 In your proof you can always choose $x\in N_0\subseteq A$ in the intersection $N\cap A$. 

