How can I improve the proof of this statement with convexity? $A\in\mathbb{R}^{m\times n}$. How do I prove that if $S\subseteq \mathbb{R}^m$ is
  convex then so is $A^{-1}(S) =\{ x \in \mathbb{R}^n : Ax
  \in S\}$ ?
My proof:
$\forall t \in [0,1], ta+(1-t)b \in S$
Since $Ax\in S$, define $Aa'\in S, Ab'\in S$,
$\forall t \in [0,1], tAa'+(1-t)Ab' \in S$
$A(ta'+(1-t)b') \in S$
$A^{+}A(ta'+(1-t)b')= ta'+(1-t)b'\in A^{+}S$   (1), where $A^{+}$ is the pseudo-inverse of matrix $A$
Since $a',b' \in \{x \in \mathbb{R}^n,Ax\in S\}$ and $x \in A^{+}S$,
we have $a',b' \in A^{+}S$   (2).
I have my proof by timing $A^+$ (the pseudo-inverse of A) before $Ax$, however that's not a common method, moreover, $A^{+}A$ is not guaranteed to be an identity matrix, which may be problematic. Is there any more common technique in proving this statement?

I think you're overcomplicating a touch,
let $x,y\in A^{-1}(S)$ then if
$$tx + (1-t)y \in A^{-1}(S)$$ for all $t\in [0,1]$, the set is convex. 
Now, let $z = tx+(1-t)y$, then $z\in A^{-1}(S)$ iff $Az \in S$. You can now write:
$$Az = ta + (1-t)b$$ 
letting $a=Ax, b=Ay$, both $a,b\in S$ since $x,y\in A^{-1}(S)$, by convexity $ta+(1-t)b\in S$ and therefore $z\in A^{-1}(S)$.
