The elements of the centralizer of $A$ commute with all elements of $A\subseteq G$ and using this to prove that $C_G(A)$ includes inverses.

I just want to make sure this is correct.

In what follows, $ G$ is a group under multiplication.

(1) For all $g\in G$ if $g$ belongs to the centralizer of $A\subseteq G$ , we have: $ ga= ag $ , for all $a\in A$.

Saying that $g\in C_G(A)$ means ( by definition ) that , $\forall a\in A$ , $gag^{-1} = a$.

Now $ gag^{-1} = a \implies gag^{-1}g = ag \implies ga = ag$.

(2) Using $ ga=ag$ to prove the existence of inverses in $C_G (A)$ :

Let $g^{-1}$ be the inverse of $g$, an arbitrary element of $C_G(A)$.

$ g^{-1} \in C_G(A) \iff g^{-1}a (g^{-1})^{-1} = a \iff g^{-1}ag =a \iff g^{-1}ga =a \iff ea=a \iff a=a$.

Since $g^{-1} \in C_G(A)$ is equivalent to a tautology ( namely $ a=a$) , $g^{-1} \in C_G(A)$ is true, for all $g\in C_G(A)$

It is a bit unclear as to what exactly you are trying to prove. By definition if $A\subset G$, the centralizer of $A$ in $G$, $C_G(A)=\{g\in G: ga=ag \text{ for all } a\in A\}$. I am unsure what you mean by "trying to prove the existence of inverses in $C_G(A)$". I think the more formal statement would be if $g\in C_G(A)$, then $g^{-1}\in C_G(A)$, and that will be true as follows. If $g\in C_G(A)$, then $ga=ag$, so multiplying this equation on the left by $g^{-1}$ and the right by $g^{-1}$ gives us that $ag^{-1}=g^{-1}a$, so we conclude that $g^{-1}\in C_G(A)$.

In fact you can show that $C_G(A)$ is a subgroup since the identity commutes with every element of $G$ (and hence $A$).