We know that the definition of local max value is "$f(a)$ is called $\textbf{local max value of f} $ if $f(a)\geq f(x)$ for all $x$ in some open interval containing $a$."
Would it be okay if I change the word "open" with "closed" i.e some closed interval?
On
Changing the definition will make some changes. Consider the function $f(x)=\frac{1}{x}$ for all $x\in (0, \infty)$. As per our original definition, $f$ does not have any local maximum. But suppose we replace our open interval by closed interval then, at every point, $a \in (0,\infty)$, $f$ has a local maximum. I suggest you verify this claim! We consider singleton sets as closed intervals, which will make every point local maximum point. You can refer to this question.
To understand the importance of considering the open intervals, I recommend you to go through the discussions in this question. Why are neighborhoods assumed to be open?
On
The other two answers point out that since $[a,a]$ is a closed interval, this definition makes every point into a local max or min.
This might feel like a technicality. You might be more interested in a definition like this one:
$f(a)$ is called a local maximum if $f(a) \geq f(x)$ for all $x$ in some closed interval $[c,d]$ with $a \in [c,d]$ and $c < d$.
This also doesn't work. Consider $f(x) = x$. Then for every number $a$, $f(a)$ is a local max. Use the interval $[a-1,a]$ to see this.
Now you might object that you don't want to allow $a$ to be one of the endpoints. So your modified definition would be
$f(a)$ is called a local maximum if $f(a) \geq f(x)$ for all $x$ in some closed interval $[c,d]$ with $a \in [c,d]$ and $c < d$ and $a\neq c$ and $a \neq d$.
However this last definition is equivalent to the standard definition. If $a$ cannot be an endpoint, you might as well rule them out by considering an open interval instead.
Your proposal would not give an equivalent definition because the singleton $\{a\}$ is a "closed interval containing $a$".