Prove a one sided inverse matrix $B_{m\times m}$ of $A_{m\times m}$ is two sided inverse. I am trying to proof that if $A_{m\times m}$ has inverse $B_{m\times m}$ then  $A_{m\times m}B_{m\times m}=Id_{m\times m}\leftrightarrow B_{m\times m}A_{m\times m}=Id_{m\times m}$.
Proof:
Suppose $V,W$ are two finite dimensional vector spaces with bases $A=\{a_1,...,a_m\},B=\{b_1,...,b_m\}$ respectively, $\forall T\in \mathrm{Hom}_{\mathbb{F}}(V,W)$ , $A_{m\times m}$ is defined as $M_{A\rightarrow B}(T)$. Since $B_{m\times m}$ exists then by definition $M_{B\rightarrow A}(T^{-1})$ exists. Therefore we have:
$A_{m\times m}B_{m\times m}=M_{A\rightarrow B}(T)M_{B\rightarrow A}(T^{-1})=M_{B\rightarrow B}(T\circ T^{-1})=Id_{m\times m}=M_{A\rightarrow A}( T^{-1}\circ T)=M_{B\rightarrow A}(T^{-1})M_{A\rightarrow B}(T)=B_{m\times m}A_{m\times m}$ 
as required. Is there any logical problem with the last step?
 A: if they are square matrices then $A$ and $B$ can be broken down into product of elementary matrices, i.e. $$A=E_1E_2...E_k$$
$$B=E_k^{-1}E_{k-1}^{-1}...E_1^{-1}$$
One can check $AB=BA=I$
A: Your proof is transferring the corresponding result from linear transformations to matrices (and your gap in logic was going from the assumption that $T$ had a right inverse to the conclusion that the right inverse was also a left inverse), and so we want to establish that result first.  If you already have that at your disposal, then all you need to do is cite the result for linear transforms.  If not, here is a proof.
Theorem: If $V,W$ are vector spaces of dimension $n$, and $S:V\to W$, $T:W\to V$ such that $S\circ T=\operatorname{Id}_W$, then $T\circ S=\operatorname{Id}_V$.  
We will need a few ingredients for the proof.
First, by the rank-nullity theorem, combined with the observation that a linear transformation is injective if and only if it has zero kernel and surjective if and only if its rank is equal to the dimension of the image space, we have that a linear map between two spaces of dimension $n$ is injective if and only if it is surjective.
Second, a bijective map of sets has a two sided inverse (or both a left inverse and a right inverse, which by Emilio's argument, must be equal).
Now, if $S\circ T=\operatorname{Id}_W$, which is a bijection, then $S$ is a surjection and $T$ is an injection, but because $S$ and $T$ are maps between equal dimensional spaces, they are both bijections.  Because bijective set maps have unique inverses, this shows that $S$ and $T$ are actually inverses of each other, and so $T\circ S=\operatorname{Id}_V$ as desired.

Here is a different perspective, related to your comment that $T\circ S \circ T = T$.  In category theory, instead  of using the notion of injection and surjection, they take the cancelability properties of injective and surjective maps as their replacements.  In particular, if $f$ is injective, and $fg(x)=fh(x)$, then $g(x)=h(x)$, and since this holds for every $x$, this implies that $g=h$. We say that injective maps are left-cancelable, or "monomorphisms" (monic for short).  Similarly, if $f$ is surjective, then $gf=hf$ implies that $g=h$.  We say that surjective maps are right-cancelable, or "epimorphisms" (epic for short).  It is an instructive exercise to show that epic set maps are surjective and monic set maps are injective.
So assume that $ST=I$. Then $TST-T=0$, which we can factor as $(T\circ S-I)\circ T=0\circ T$, and we would like to be able to cancel the $T$'s on the right.  From the previous solution, we have that $S\circ T=I$ implies that $T$ is injective, rank-nullity has that $T$ is surjective, and since surjective maps are epic, we can cancel on the right to get $TS-I=0$ or $TS=I$.
This is essentially the proof you gave, except placed into a broader context.
A: If:
$$
AB=I \quad \mbox{and} \quad CA=I
$$
than
$$
(CA)B=IB \Rightarrow C(AB)=B \Rightarrow CI=B \Rightarrow C=B 
$$
