I am reading "Supplement for Measure, Integration & Real Analysis" by Sheldon Axler.
This text is freely available from the author's site.
The following sentences are in this text.
suppose $G$ is an open subset of $\mathbb{R}$. For each $t\in G$, let $G_t$ be the union of all the open intervals contained in $G$ that contain $t$. A moment’s thought shows that $G_t$ is the largest open interval contained in $G$ that contains $t$.
My proof:
Since $G_t$ is the union of all the open intervals contained in $G\subset\mathbb{R}$, $G_t$ is an open subset of $\mathbb{R}$.
Let $x,y\in G_t$ such that $x<y$.
Let $z\in (x,y)$.
Since $G_t$ is the union of all the open intervals contained in $G\subset\mathbb{R}$ that contain $t$, $x\in (a,b)$ for some $(a,b)$ which is an open interval contained in $G\subset\mathbb{R}$ that contains $t$ and $y\in (c,d)$ for some $(c,d)$ which is an open interval contained in $G\subset\mathbb{R}$ that contains $t$.
(Case 1):
If $z<b$, then $a<x<z<b$.
So, $z\in (a, b)$.
(Case 2):
If $b\leq z$, then $c<t<b\leq z<y<d$.
So, $z\in (c, d)$.
So, $z\in (a, b)\cup (c,d)\subset G_t$.
Therefore $G_t$ is an interval.
Let $I$ be any open interval contained in $G$ that contains $t$.
Then, $I\subset G_t$ by the definition of $G_t$.
$\therefore$ $G_t$ is the largest open interval contained in $G$ that contains $t$.
Is my proof ok?
Your proof seems ok. It could be simpler in case you know the following general topological nonsense: if $(A_i)_{i\in I}$ are connected subsets of the topological space $X$, and $A_i\cap A_j\neq\emptyset$ for any $i,j\in I$, then $\bigcup_{i\in I}A_i$ is connected (actually weaker condition suffices).
To apply that in this situation, a subset of $\mathbb{R}$ is connected iff it's an interval, and an open subset is connected iff it's an open interval. Each member of $G_t$ contains $t$, so of course they pairwise intersect, and the union is connected, hence an open interval.