Let $H$ be the centralizer of $a_0$.  $H$ certainly contains the centralizer $MA$ of $A$.  We now compare their Lie algebras, $\mathfrak{h}$  and $\mathfrak{m}+\mathfrak{a}$.  Since $H$ contains $A$ it is normalized by it and we can decompose $\frak{h}$ under this action.  Now $\frak{m}+\frak{a}=\frak{g}_0\subset\frak{h}$ is the subspace corresponding to the trivial character of $A$, so the rest of $\frak{h}$ must transform under other characters (that is, under the roots).  However, no root vanishes on $a_0$ (since $a_0$ is not on any wall), so no root can occur in this action.  It follows that $\frak{h} = \frak{a}+\frak{m}$.

This shows that the connected components of $H$ and $MA$ agree.  In the algebraic category we are done (the centralizers of a semisimple element and of a torus are both connected).  I can't recall the argument in the smooth category off the top of my head.