Matrices with n distinct real eigenvalues

Idea. $\mathcal{D}$ is the image of $GL_n^+\times\mathcal{C}$, with $\mathcal{C}$ the convex cone of strictly increasing diagonals, under a continuous map.

Step 1: parametrisation. Let $A\in\mathcal{D}$ with eigenvalues $\lambda_1<\dots<\lambda_n$. The eigenvectors $v_i$ ($Av_i=\lambda_iv_i$) are linearly independent (eigenvectors for distinct eigenvalues), so $P=(v_1|\dots|v_n)\in GL_n(\mathbb{R})$ and
$$A=P\operatorname{diag}(\lambda_1,\dots,\lambda_n)P^{-1}.$$

Step 2: $\det P>0$ can be arranged. Replacing $v_1$ by $-v_1$ changes $\det P$ to $-\det P$ and does not change $A$ (equivalently, $P\mapsto PD'$ with $D'=\operatorname{diag}(-1,1,\dots,1)$ commutes with the diagonal matrix). So we may take $P\in GL_n^+$.

Step 3: the parameter space. $\mathcal{C}=\{(\lambda_1,\dots,\lambda_n):\lambda_1<\dots<\lambda_n\}\subset\mathbb{R}^n$ is convex (an intersection of open half-spaces), hence path connected. $GL_n^+$ is path connected (Problem 8). So $GL_n^+\times\mathcal{C}$ is path connected.

Step 4: the map. $\Theta:GL_n^+\times\mathcal{C}\to M_n(\mathbb{R})$, $\Theta(P,\lambda)=P\operatorname{diag}(\lambda)P^{-1}$ is continuous (the inverse is a rational function with denominator $\det P\ne0$), and by Steps 1 and 2 its image is exactly $\mathcal{D}$. Therefore $\mathcal{D}$ is the continuous image of a path connected set.

Conclusion. $\mathcal{D}$ is path connected.

Remark. The Weyl chamber $\mathcal{C}$ (ordering the eigenvalues) is what removes the ambiguity of permuting the eigenvectors. The proof shows more: $\mathcal{D}$ is an open dense subset of the set of matrices with real spectrum. By contrast the set of matrices with $n$ distinct (possibly complex) eigenvalues is also connected when $n\ge2$ but requires a different argument.