If $x$ is an isolated point of $S \subseteq \mathbb{R}$, then $x$ is a boundary point of $S$. Is the following proof valid? (Note: I know there is a post discussing this problem, but I am curious to see if my argument works). This problem is different from another post that is similar with this one, because my proof begins with an existential neighborhood and ultimately considers a different route in proving the problem.
Let $S \subseteq \mathbb{R}$. Prove that if $x$ is an isolated point of $S$, then $x$ is a boundary point of $S$.
Suppose that $x$ is an isolated point of $S$ and assume to the contrary that $x$ is not a boundary point of $S$. Then for each $x$, there exists an $\epsilon > 0$ such that $N_{\epsilon} \subseteq S$. But because $x$ is not an accumulation point of $S$, we see that for each $x$, there exists a $\delta > 0$ such that $N_{\delta}^* \cap S = \varnothing$ where $N_{\delta}^*$ is some deleted neighborhood. Therefore, $N_{\delta}^* \subseteq S^c$. 
Thus there are two intervals centered at $x$, and either $\epsilon \geq \delta$ or $\delta \geq \epsilon$, which would then imply that $N_{\delta}^* \subseteq N_{\epsilon}$ for the former and $N_{\epsilon} \subseteq N_{\delta}^*$ for the latter (it helps to draw a picture of these intervals to see why these cases arise). But in both cases we reach a contradiction, because these possibilities imply that $N \not \subseteq S$. Therefore, there are no $N_{\epsilon}$'s contained in $S$, but because $x \in S$, we deduce that every neighborhood $N$ must satisfy $N \cap S \neq \varnothing$ and $N \cap S^c \neq \varnothing$. Hence, $x$ is a boundary point of $S$.
 A: Note that $x\in N_\epsilon$ and $x\notin N_\delta^*,$ so we can't have $N_\epsilon\subseteq N_\delta^*.$ Rather, we can conclude that $N_\delta\subseteq N_\epsilon$ or $N_\epsilon\subseteq N_\delta.$
I would recommend proving your claims about $N_\epsilon$ and $N_\delta$ explicitly, rather than asking your reader to draw a picture. For one thing, one can be misled by pictures, and for another, it helps you to make certain you're on the right track with your proof. It helps to consider $$N:=N_\epsilon\cap N_\delta=N_{\min(\epsilon,\delta)},$$ and use the assumptions to show that $N\subseteq S$ and $N^*\cap S=\emptyset,$ so that $N=\{x\},$ which is absurd.
Alternately, the simplest proof is not by contradiction, but direct!

Suppose $x$ is an isolated point of $S,$ so that there is some $\delta>0$ such that $N_\delta(x)\cap S=\{x\}.$ Take any neighborhood $U$ of $x.$ By definition, there is some $\epsilon>0$ such that $N_\epsilon(x)\subseteq U.$ Put $c=\min\{\delta,\epsilon\},$ so $c>0.$ Put $y=x+\frac12c.$ Then $y\ne x$ and $|y-x|<\delta,$ so $y\notin S.$ But $|y-x|<\epsilon,$ so $y\in U.$ Thus, $S^c\cap U\ne\emptyset,$ and trivially $S\cap U\ne\emptyset,$ and so $x$ is a boundary point of $S.$

