real analysis - Every point in open set belongs to component interval

I'm currently self studying analysis, and I'm reading Apostol's Mathematical Analysis, which proves the following statement about component intervals.

Definition of component interval: Let $S$ be an open subset of $\mathbf{R}^1$, An open interval $I \subseteq S$ is a component interval if there does not exist an open interval $J\ne I$ such that $I \subseteq J \subseteq S$.

Statement. Prove that every point of a nonempty open set S belongs to one and only one component interval of $S$.

Proof. Assume $x \in S$. then $x$ is contained in some open interval $I$ with $I \subseteq S$. There are many such intervals but the "largest" of these will be the desired component interval. We leave it to the reader to verify that the largest interval is $(a(x), b(x))$, where $a(x) =$ inf $\{ a: (a,x) \subseteq S\}$ and $b(x) =$ sup $\{ b: (x,b) \subseteq S\}$. (The proof continues...)

This is intuitively true, since any interval greater than this obviously contains points that are not in $S$. I've tried defining such a point but am unable to show that it doesn't lie in $S$. Can someone prove this for me?

Answer

You want to show that for all $\epsilon>0$, neither $I_{\epsilon}:=(a(x)-\epsilon,b(x))$ nor $J_{\epsilon}:=(a(x),b(x)+\epsilon)$ are contained in $S$.

Since $\epsilon>0$, there is a point $y\in\mathbb{R}$ with $a(x)-\epsilon

Similarly, one can show that $J_{\epsilon}$ is not a subset of $S$.