Is There a Problem with This Professor's Proof Concerning Interior and Boundary Points? Here is a professor's solution to the exercise which states, " Prove that if $x$ is an isolated point of a set $S \subseteq \mathbb{R}$, then $x$ is a boundary point of $S$."

The professor derived a contradiction upon assuming that for an arbitrary neighborhood $N$ centered at an isolated point $x$ of $S$, we have $N \subseteq S$, so he then proceeds in saying that every $N$ must have a noenmpty intersection with $S^c$. But I feel that there is something wrong with his argument: Using mathematical logic, we can represent his assumption that every $N$ is a subset of $S$ as $\forall \, N : p(N)$ where $p(N)$ means, "$N$ is contained in $S$." He then showed that this assumption is false, so $\sim \forall \, N : p(N) \equiv \exists \, N : \, \sim p(N)$ is true, which means there are some $N$'s that are not contained in $S$. But this information is not sufficient in concluding as to whether or not there is at least one $N$ contained in $S$; it just means not every $N$ is contained in $S$. But if I am right, how can he proceed with his proof since whatever result he deduces about $N$ after the contradiction does not necessarily apply to all $N$?
I hope what I am trying to convey is clear, and I thank you for your help

Note that a contradiction is reached for any particular neighborhood $N\subseteq S$.
Hence, for any neighborhood $N$ you do have $N\cap(\mathbb R\setminus S)\not=\emptyset$. This is what is used later on.
