Prove that if X is a metrizable topological space and Y is homeomorphic to X, then Y is metrizable
Since X is a metrizable topological space, we have a metric space (X, d).
Let Y be a topological space homeomorphic to X and $\displaystyle f:X \rightarrow Y$ be a homeomorphism.
Define d' on $\displaystyle Y \times Y$ such that
$\displaystyle d'(y_1, y_2) = d( f^{-1}(y_1), f^{-1}(y_2)), y_1, y_2 \in Y$.
I'll leave it to check d' is indeed a metric.
Since both $\displaystyle f$ and $\displaystyle f^{-1}$ are continuous bijection, we see that $\displaystyle f$ and $\displaystyle f^{-1}$ are isometries, which implies that an open ball of radius r >0 with respect to a metric d on space X corresponds to an open ball of radius r with respect to a metric d' on space Y, and vice versa. Now, the open balls in Y defined by d' can be given as a basis for a topological space Y. Thus, Y is metrizable.