How to prove that bijective functions have surjective inverses I am trying to prove

Since $f$ is bijective therefore it has an inverse and $f^{-1}$ is
  bijective

But stuck on proving that $f^{-1}$ is surjective
Suppose $f: A \to B$ is a bijection, then a proof of $f^{-1}$ is surjective is offered in another post which says

If $x \in A$ then $f(x) \in f(A) = B$ then $x \in f^{-1}(B)$.  If $y =
> f^{-1}(B)$ then there exists a $f(y) \in B$ so that $f^{-1}(f(y)) = y
 \in A$.  So $f^{-1}(B) = A$.

I have no idea what the proof is trying to do. Can someone construct a more lucid version of the proof or clarify what the proof is trying to achieve? 

For all $x\in A$ there is $y\in B$ such that $f(x)=y$; thus $f^{-1}(y)=x$. Therefore $f^{-1}$ is surjective.
