$\mathrm{GL}_n(\mathbb{R})$ does not contain matrices with determinant zero. This leaves some matrices with positive determinants and some with negative determinants. If these are each connected open sets, then they form a separation of $\mathrm{GL}_n(\mathbb{R})$, so the whole thing cannot be connected.
Consider that any matrix with positive determinant can be converted to one with negative determinant by multiplying the first row (or column) by $-1$, giving an easy way to show a bijection. (Showing it's a homeomorphism is only a bit more work.)
Open... Let $P = \{M \mid \det M > 0\}$ and consider $f:P \rightarrow \mathbb{R}:M \rightarrow \det M$. If you believe this is a continuous function then we observe that $f(P) = (0, \infty)$, which is open, so the preimage, $P$, is open. To show that it is continuous, (I abbreviate:) small changes in the matrix entries cause small changes in the determinant.