Suppose group $G$ is generated by $\{g_1,\dots g_n\}$. Let $A$ be nonempty subset of $G$, and $C_G(A)$ is centralizer of $A$ in $G$. If for some $1<m<n$, then we have that $g_1, g_2,\dots,g_m\in C_G(A)$, but $g_{m+1},\dots,g_n\notin C_G(A)$. I can see that, then the elenents of $G$, that are generated only by a subset of $g_1,g_2,\dots, g_m$ are in $C_G(A)$, but if $g\in G$ is not purely generated by a subset of a subset of $g_1,g_2,\dots, g_m$, then $g\notin C_G(A)$.
I think the second assertion follows, because, since $C_G(A)$ subgroup, and if, e.g., $g=g_1\cdot g_{m+1}\cdot g_n\in C_G(A)$, then $g_1^{-1}\cdot g_1\cdot g_{m+1}\cdot g_n=g_{m+1}\cdot g_n\in C_G(A)$. I first thought this gives contradiction. But, now I think, even if $g_{m+1}, g_n\notin C_G(A)$, what forces us to conclude $g_{m+1}\cdot g_n\notin C_G(A)$?
You are right that any element of the subgroup generated by $g_1,\ldots,g_m$ is in $C_G(A)$, but not that any element not in that subgroup is not in $C_G(A)$.
For a counter-example, consider the subgroup $G$ of $S_7$ generated by $g_1=(1,2,3)$, $g_2=(2,3)(4,5)$, $g_3=(2,3)(6,7)$ and $A=\{g_1\}$. Then you can check that $g_1\in C_G(A)$, $g_2,g_3\notin C_G(A)$ but $g_2\cdot g_3\in C_G(A)$.