Components of the set where the trace is non-zero

Answer. Exactly two.

Idea. The set is the union of two open half-spaces, each convex, hence connected.

Step 1: the two pieces. Let $H_+=\{\operatorname{tr}>0\}$ and $H_-=\{\operatorname{tr}<0\}$. They are open (preimages of open sets under the continuous map $\operatorname{tr}$), disjoint, non-empty (they contain $I$ and $-I$ since $\operatorname{tr}(\pm I)=\pm n$), and their union is the set in question.

Step 2: each piece is connected. $\operatorname{tr}$ is linear, so if $\operatorname{tr}A>0$ and $\operatorname{tr}B>0$ then $\operatorname{tr}((1-t)A+tB)=(1-t)\operatorname{tr}A+t\operatorname{tr}B>0$. So $H_+$ is convex, hence path connected (Problem 1); likewise $H_-$.

Step 3: conclusion. The set is a disjoint union of two non-empty open connected sets, so these are precisely its two components.

Remark. The same argument applies to the complement of any hyperplane: $\{\varphi\ne0\}$ for a non-zero linear form $\varphi$ always has two components. The complement of a subspace of codimension $\ge2$, on the other hand, is connected (Problem 13).