I am reading a theorem that says
A topological space is regular if and only if $\forall x \in X$ and for all $u\in T$ such that $x\in u$, there exists $v\in T$ such that $x\in v\subseteq \bar{v}\subseteq u$.
The first few lines of the proof are as follows
Let $(X, T)$ be a regular space and let $x\in X$ and $u\in T$ such that $x\in u$
Let $A=X-u$, $x\notin A$
Then there exists $v,w\in T$ such that $x\in v$, $A\subseteq w$ and $v\,\cap\,w=\phi$ $\Longrightarrow x\in v\in \bar{v}$
$v\subseteq X-w\subseteq X-A=u$
The remaining lines are to prove that $\bar{v}\subseteq{u}$.
My question is: I am confused because this theorem seems to suggest that in a regular space $(X, T)$ there are an infinite number of open subsets in $T$ that contain $x$ because for every $x\in u\in T$ there is $v\in T$ such that $x\in \bar{v}\subseteq u\in T$ but there are many regular spaces where $T$ contains a finite number of sets?
For example $X=\{a,b,c\}, T=\{\phi,X,\{a\},\{b,c\}\}$ is regular space. But $a\in u=\{a\}\in T$ and there is no $v\in T$ such that $x\in v\subseteq \bar{v}\subseteq u=\{a\}\in T$.
In your example $v = u$ will do. Sets can be closed and open at the same time (often called clopen). So $v \subseteq \overline{u}$ need not imply $ u \neq v$ at all.