Continuous bijection between compact and Hausdorff spaces is a homeomorphism

The continuous image of a compact space is compact. We don't need sequences to see this; in fact sequences don't even suffice to see it, in general. The definition of compactness is by open covers, so use that:

If $f:X \to Y$ is continuous, $A \subseteq X$ is compact, then consider an open cover $O_i, i \in I$ of $f[A]$. Then $f^{-1}[O_i], i \in I$ is a cover of $A$ (by basic set theory) and an open cover as $f$ is continuous. So finitely many $f^{-1}[O_i], i \in F$ (so $F \subseteq I$ finite) exist that also cover $A$ and again simple set theory tells us that the $O_i, i \in F$ is a finite subcover of the original cover for $f[A]$. Hence $f[A]$ is compact.

The lemma then follows from the basic fact that if $Y$ is Hausdorff, and $B \subseteq Y$ is compact, then $B$ is closed in $Y$. This too is shown using open covers and the definition of Hausdorffness. Plenty of proofs can be found online.

Now if a bijection $f: X \to Y$ is closed, this is the same as saying its inverse map $g: Y \to X$ is continuous: $g$ is continuous iff $g^{-1}[C]$ is closed for all closed $C \subseteq X$. And $g^{-1}[C] = f[C]$ because $g$ is the inverse of the bijection $f$. As $f$ is a closed map by the lemma, you're done.