Proving $\partial \bar N\subset \partial N$ I wanted to prove $\partial \bar N\subset \partial N$ in a different way, by showing if $x \in \partial \overline{N}$ then $x \in \partial N$. This I wanted to show by showing that if $U$ is an nbhood of $x$ then it contains 
(i) a point of $N$ and 
(ii) a point of $N^c$. 
Since $U$ is an nbhood of $x$ it contains a point of $\overline{N}$ and a point of $\overline{N}^c$. We have $N \subset \overline{N}$ hence $\overline{N}^c \subset N^c$ hence (ii) is clear. What I'm stuck with is (i). I tried the following: $U$ contains a point $y$ of $\overline{N} = \partial N \cup \mathrm{int}N$. If $y \in \mathrm{int}N$ then we're done. If $y$ in $\partial N$ then either $y \in N$ or $y\in N^c$. If $y \in N$ we're done.
But if $y \in N^c$ then... nothing. So it looks as if $U$ could be a subset of $N^c$. Is it really not possible to prove it like this or do I just not see how? Thanks for help.
 A: Clarification: In this post whenever I say the word neighborhood I mean an open neighborhood. (Based on the comments, it seems that this caused some misunderstandings. And I have to admit that I did not think about this when posting.)
If $x\in\overline N$ if and only if every neighborhood $U$ of $x$ contains a point from $N$. 
$\boxed{\Rightarrow}$ Suppose that there is a neighborhood $U$ such that $U\cap N=\emptyset$. Then $N\subseteq X\setminus U$ and $X\setminus U$ is closed, hence 
$\overline N\subseteq X\setminus U$. I.e., $\overline N\cap U=\emptyset$, contradicting the assumption that $x$ is in this intersection.
$\boxed{\Leftarrow}$ If $x\notin\overline N$ then $U:=X\setminus\overline N$ is a neighborhood of $x$ such that $U\cap N=\emptyset$.

From this you can easily see that if every neighborhood of $x$ contains a point from $\overline N$, then every neighborhood of $x$ contains a point from $N$.
(Let $U$ be a neighborhood of $x$. If $U\cap\overline N\ne\emptyset$, then we have a point $y\in U\cap \overline N$. Since $U$ is a neighborhood of $y$ and $y\in\overline N$, we get that $U\cap N\ne\emptyset.$)
