A continuous bijective function from $X$ to $Y$ is a homeomorphism Theorem: Let $f:X\rightarrow Y$ be a continuous bijection, $X$ is locally compact, and $Y$ is Hausdorff. Then $f$ is a homeomorphism.
So let $U\subset X$ be closed, it's enough to show $f(U)$ is closed in $Y$.
Since $X$ is locally compact, for each $x\in U$, there exists a compact neighborhood $U_x$, and $U_x\cap U$ is also a neighborhood of $x$. Now $U=U\cap(\cup_{x\in U}U_x)$ where each $U_x$ is a compact subset of $X$. But how do I show that $f(U)$ is compact in $Y$?
Now suppose that I've shown that $f(U)$ is compact in the Hausdorff space $Y$, I can show that for each $y$ not in $f(U)$, there exists an open neighborhood of $y$ that does not intersect $f(U)$. But how does that lead to $f(U)$ being closed?

The theorem you are trying to prove is false. Consider $X=[0,1)$, $Y=S^1$ and $f(x)=(\cos (2\pi x),\sin (2\pi x))$. This is a continuous bijection of locally compact metric spaces, but it is not a homeomorphism.
As a fact to keep in mind, it is true that a continuous bijective map $\Bbb R^m\to \Bbb R^m$ is a homeomorphism, but it requires more than general topology to be proved.
Added: The same theorem with "$X$ compact" instead of "locally compact" is true, and perhaps this was the intention of the assignment.
