I wanted to prove equivalence of norms in finite dimensional normed spaces without using Heine-Borel in arbitrary spaces.
Let $(E, ||.||)$ be a finite dimensional normed space. In the process of proving every finite dimensional normed space is complete, I was able to construct a linear homeomorphism $T: \mathbb{R}^n \to E$ which is just the mapping from a vector of coordinates to the element of $E$ with said coordinates in a finite basis of $E$.
I then tried extending this result to prove the equivalence of any two norms on $E$, but it turned out to be rather simple and makes me question whether I'm missing something:
Let $||.||_a$, $||.||_b$ be two norms on $E$. By the previous comment, I know there exist two linear homeomorphisms: $$T_a : (\mathbb{R}^n, ||.||_2) \to (E, ||.||_a)$$ $$T_b : (\mathbb{R}^n, ||.||_2) \to (E, ||.||_b)$$ And I can write the identity function as a composition of these two: $$T_b \circ T_a^{-1} = id: (E, ||.||_a) \to (E, ||.||_b)$$ Since the composition two homeomorphisms is again an homeomorphism, it follows that the identity function is a homeomorphism, and that $||.||_a$ and $||.||_b$ are equivalent.
Is my proof correct? Any thoughts/comments are greatly appreciated, thanks!