I'm reading Apostol's book. This definition and theorem come from it
Definition: Let $S$ be an open subset of $R^1$. An open interval $I$ (which may be finite or infinite) is called a component interval of $S$ if $I \subseteq S$ and if there is no open interval $J \neq I$ such that $I \subseteq J \subseteq S$
Theorem: Every point of a nonempty open set $S$ belongs to one and only one component interval of $S$.
I will now give a proof of this theorem. Can someone verify that everything I did is correct? If necessary, I can include Apostol's proof as well. Just leave a comment.
Proof: Let $x \in S$. Define $I_x:=(a(x),b(x))$, where $a(x) = \inf A$ and $b(x) = \sup B$ with $A := \{a < x : (a,x) \subseteq S\}$ and $B := \{b > x: (x,b) \subseteq S\}$
and where $a(x), b(x)$ might be $\pm\infty$ (i.e. $A,B$ are unbounded). Of course, one must see that the sets $A,B$ are non empty, to conclude the existence of $a(x)$ and $b(x)$, but this is alright, because $S$ is open, and hence there is an interval $(x-\epsilon, x + \epsilon) \subseteq S$
We now prove that $I_x \subseteq S$. Let $y \in I_x$. WLOG, assume that $x \leq y$. Take $z \in (y,b(x))$ satisfying $(x,z) \subseteq S$ (such $z$ exists, because otherwise for any $z \in (y,b(x))$, we would have that $(x,z) \not\subseteq S$, which is not possible, by definition of $b(x)$)
It now follows that $y \in (x,z) \subseteq S$.
Now, let $J$ be an open interval such that $I_x \subseteq J \subseteq S$. Then we can write $J = (s,t)$ where $s,t$ might be $\pm \infty$.
But $x \in I_x$, so $x \in J$ and hence $(s,x) \subseteq J \subseteq S$, so $s \in A$ and $a(x) \leq s$. In the same way, $b(x) \geq t$, so we conclude that $J = (s,t) \subseteq (a(x),b(x)) = I_x$. This proves that $I_x$ is a component interval.
For the uniqueness, assume that $I,J$ are component intervals containing $x$.
We then have:
$I \subseteq I \cup J \subseteq S$ and $J \subseteq I \cup J \subseteq S$. Because $I \cup J$ is an open interval, the definition of component interval yields that $I = I \cup J = J$, and the uniqueness follows.
What you did seems correct. I would simplify considering
$$\mathcal I_x =\{I ; I \text{ is an open interval containing } x \text{ included in } S\}$$
Then $\bigcup \mathcal I_x$ is open as a union of open subsets, included in $S$ as all its elements are included in $S$ and is an interval as for all $I \in \mathcal I_x$, $x \in I$. A union of connected subsets all containing a point $x$ is connected.
Moreover $\bigcup \mathcal I_x$ contains all the open intervals containing $x$ by definition. So it is a component interval of $S$.
Unicity follows from the fact that an open interval containing $x$ and included in $S$ is included in $\bigcup \mathcal I_x$.