Conjecturing concerning $\mathrm{cl}(\mathrm{int}S) = S$ Let $S$ be a subset of $\mathbb{R}$, $\mathrm{int}S$ denote the set of interior points of $S$, $\mathrm{bd}S$ denote the set of boundary points of $S$, $S'$ denote the set of accumulation points of $S$, and $\mathrm{cl}S$ denote the closure of $S$.
The problem states that

Provide a counterexample to the following claim: If $S$ is a closed set, then $\mathrm{cl}(\mathrm{int S}) = S$.

An easy counterexample is to consider a nonempty, finite subset $S$. The $S$ is closed because $S' = \varnothing \subseteq S$. But because $\mathrm{int}S = \varnothing$, we have $\mathrm{cl}(\mathrm{int}S) = \varnothing \neq S$.
My Question
So I have a conjecture and an attempt to proving it: If $S$ is a closed, infinite set, then $\mathrm{cl}(\mathrm{int}S) = S$. I was wondering if the following proof is valid.
Suppose that $S$ is a closed, infinite set. Without loss of generality, let $a, b \in \mathbb{R}$ such that $a < b$ and $S = [a,b]$.  Then
$$\begin{align} \mathrm{cl}(\mathrm{int} S) 
     &= \mathrm{int}S \cup \mathrm{bd}(\mathrm{int}(S)) \\
     &= (a,b) \cup \{a,b\} \\
     &= S
\end{align}$$
as desired.

You cannot assume that $S$ is an interval. There are infinite closed sets with empty interior; one interesting example is the Cantor set.
