$\partial A $ is a closed set Attempt:
Pick a point $x \notin \partial A$, and Take a nghbd $N$ of $x$. If $N \cap \partial A = \varnothing$, then we are done.
Suppose $y \in N \cap \partial A$. So, $y \in \partial A$, therefore $N \cap A = \varnothing$. But, since $N$ is a neighborhood of $x$, $x \notin \partial A$. So, we must have $N \subseteq A$. In other words, $N$ cannot contain points not in $A$. But, since $y \in N$ and $y \in \partial A$, $N$ must contain at least one point not in $A$. Contradiction, therefore, $N \cap \partial A = \varnothing$ and, $\partial A$ is a closed set.
Is this proof correct? thanks for any feedback.
 A: Your proof doesn't work. In particular, it falls apart when you conclude that $N\subseteq A$. For example, if we take $N$ to be the whole space, then it is a neighborhood of $x$ with non-empty intersection with $\partial A$ if $\partial A$ is non-empty, but cannot be a subset of $A$. (Also, "$y\in\partial A$, therefore $N\cap A=\varnothing$" is incorrect, but that may just have been a transcription error.)
Start over from the assumption that $x\notin\partial A.$ Since $\partial A$ is defined as the set of points $z$ such that every open neighborhood of $z$ contains a point in $A$ and a point not in $A$, this means that $x$ has an open neighborhood $N$ that is either a subset of $A$ or is disjoint from $A$. Note then that no point of $N$ can be in $\partial A$ in either case. Since this holds for all $x\notin\partial A,$ then the complement of $\partial A$ is open, and so $\partial A$ is closed.
A: I gave you a proof of this in your previous question. Again, I find your constant pursuit of contradictions strange, and this practice is giving you problems. Mainly, because you seem to try to be forcing the contradiction without carefully looking at what you're claiming. I think you'll find it better to read the definitions thoroughly, understand them and seek out, when viable, direct proofs of the statements. For example:

We prove $\partial A$ is closed. Suppose $x\notin \partial A$. By definition, there exists a nbhd $N$ of $x$ disjoint from $A$ or $X\setminus A$. Now pick a point $y$ in $N$. Then by definition $N$ itself is a nbhd of $y$ disjoint from  $A$ or $X\setminus A$. This means that $y\notin \partial A$. Thus $N$ is a nbhd of $x$ disjoint from $\partial A$; and thus $\partial A$ must be closed.

