I've often seen proofs that invoke the Heine-Borel theorem to show that certain matrix groups are compact. They view the matrix groups (such as $SO(n)$ as $\mathbb{R}^{2n}$ and then go on to prove them as closed and bounded, which proves that they are compact).
Is a similar proof strategy possible when we are considering matrices using trace norms? In particular, I wish to show that a certain subclass of matrices of $GL_n(\mathbb C)$ is compact under trace norm. Is there a homeomorphism from $(GL_n(\mathbb C), \text{trace norm})$ to some $(\mathbb R^m, \text{euclidian norm})$?
No, it's not possible.
For any $m>0$, $\mathbb{R}^m$ is contractile (i.e. has the homotopy type of a point) while $GL_{n}(\mathbb{C})$ has the same homotopy type as $U_n(\mathbb{C})$ (because of the polar decomposition). But $U_n(\mathbb{C})$ isn't contractile in general.
Therefore $\mathbb{R}^m$ and $GL_{n}(\mathbb{C})$ can't be homeomorphic.