Vtyjkyuk

I have the following question: Let $f: X rightarrow Y$ be a open, continuous map and let $X' subseteq X$ be a closed subspace such that the restriction $f|_X': X' rightarrow f(X')$ is a bijection. Does this already imply that $f|_X'$ is proper?

My attempt of a proof looks like this: If $Csubseteq f(X')$ is compact and $(U_i cap X')_iin Isubseteq X'$ is an open cover of $f|_X'^-1(C)$, the family $(U_i cup (X')^c)_iin I subseteq X$ is an open cover of $f^-1(C)$. Because $f$ is open $(f(U_i cup (X')^c))_iin I subseteq Y$ is an open cover of $C$ in $Y$, so $(f(U_i cup (X')^c) cap f(X'))_iin I subseteq Y$ is an open cover of $C$ in $f(X')$. Now we can use compactness of $C$ to find a finite subcover. But I don't see how to proceed from here. Is it even possible? Can anyone help?

The claim is false. Consider $X=Bbb R$, $Y=S^1$, $X'=[0,infty)$ and $f(x)=(cos(4arctan x),sin(4arctan x))$. $f(X')=S^1$ but $left.frightrvert_X'^-1[S^1]=[0,infty)$.