some technical quibbles:

there is no reason to suppose that for an arbitrary U in X_{1}, that f(X_{1}) ∈ T_{2}.

rather, you want to pick an arbitrary U ∈ T_{1},and show that f(U) ∈ T_{2}.

(note that saying f(T_{1}) = T_{2}is actually an abuse of notation, f is a function from X_{1}to X_{2}, not a function between the respective power sets. there is, however an induced function [f] which does map between the power sets:

[f](U) = {x_{2}in X_{2}: x_{2}= f(x_{1}) for some x_{1}in U} hopefully, no confusion will arise from this, and i will use f(U) to mean [f](U)).

if f is a homeomorphism, then since U is open, and f^{-1}is a continuous function, (f^{-1})^{-1}(U) = f(U) is open in X_{2}

(here the "inner" inverse symbol means "inverse function" and the "outer" inverse symbol means "pre-image", these two sets are equal because f is bijective).

this shows that, at the very least, f(T_{1}) is a subset of T_{2}.

on the other hand, suppose V is an open set in X_{2}(that is, an element of T_{2}). since f is continuous, f^{-1}(V) (here i mean pre-image) is open in X_{1},

so V = f(f^{-1}(V)) is the image of an open set in X_{1}. this means that T_{2}is a subset of f(T_{1}).

the other implication you prove looks ok, i would just note that both f and f^{-1}are open maps, so both are continuous.