My proof goes as follows: Consider the set Cl(a,b) and an element $x\in Cl(a,b)$, then $a<x<b$. We take an open set containing $x$: $(-\infty,x]$ and $(-\infty,x]\cap(a,b)=(a,x]\implies a\leq x$.
Analogously with the open set $[x,\infty)$ we have $[x,b)\implies x\geq b$.
It follows that $x\in [a,b]$ $\blacksquare$
I feel that my proof is too simple to be correct.
As well I have to show by example that this contention is proper.
$\newcommand{\cl}{\operatorname{cl}}$You get off on the wrong foot right at the beginning: $x\in\cl(a,b)$ does not imply that $a<x<b$, and $(\leftarrow,x]$ is not in general an open set.
The easiest way to show that $\cl(a,b)\subseteq[a,b]$ is to prove the contrapositive: if $x\notin[a,b]$, then $x\notin\cl(a,b)$. If $x\notin[a,b]$, then either $x<a$, or $x>b$. If $x<a$, then $(\leftarrow,a)$ is an open nbhd of $x$ disjoint from $(a,b)$, and if $x>b$, then $(b,\to)$ is an open nbhd of $x$ disjoint from $(a,b)$, and in either case $x\notin\cl(a,b)$.