Countinous map $F:X \to Y$, $Y$ Hausdorff. If $F$ has a continuous left inverse, then it's proper. $F:X \to Y$ is a countinous map between topological spaces, $Y$ is Hausdorff. 
If $F$ has a continuous left inverse (i.e. $\exists$ continuous map $G: Y \to X\ $ s.t. $\ G \circ F = id_X$), then it's proper (preimage of compact set is compact).

My proof:
Suppose $B\subset Y$ is compact. In Hausdorff space, compact set is closed, so $B$ is also colsed.
$A:=F^{-1}(B)=F^{-1}(B\cap F(X)), B\cap F(X)=F(A). $
$G(B\cap F(X))=G \circ F(A)=A$. 
$G$ is continuous, $F(X)=G^{-1}(X)$ is closed, $B\cap F(X)$ is closed subset of compact set $B$ and is therefore compact. Continuous function maps compact set to compact set, so $A=G(B\cap F(X))$ is closed. $\Box$
Is my proof correct? Or is there any simpler proof?
Thanks for your time and patience!

Update:
As Henno Brandsma mentioned in comment, $F(X)=G^{-1}(X)$ is not correct. 
$G^{−1}[X]=\{y∈Y:G(y)∈X\}=Y$ not $F[X]$. An alternative proof showing $F(X)$ is closed is also given in answer.

I have a better idea and a generalized conclusion:
If $X$ and $Z$ are topological space, $Y$ is Hausdorff space, $f:X\to Y$ is continuous, $g:Y\to Z$ is continuous, and $g\circ f$ is proper, then $f$ is proper.
Proof:
If $K \subset Y$ is compact, then $K \subset g^{-1}(g(K))$ and $f^{-1}(K)\subset f^{-1}(g^{-1}(g(K)))＝(g\circ f)^{-1}(g(K))$. Since $g(K)$ is compact, $(g\circ f)^{-1}(g(K))$ is compact. But $f^{-1}(K)$ is closed,thus $f^{-1}(K)$ is compact.
