The 2×2 Hurwitz matrices form a path-connected set

Idea. The Hurwitz condition is open and invariant under "shrinking towards the stable point $-I$": $(1-t)A-tI$ has eigenvalues $(1-t)\lambda-t$.

Part 1. Let $\tau=\operatorname{tr}A$, $\delta=\det A$; the eigenvalues $\lambda_1,\lambda_2$ are the roots of $t^2-\tau t+\delta$, with $\lambda_1+\lambda_2=\tau$, $\lambda_1\lambda_2=\delta$.
- *Non-real roots.* They are $\lambda,\bar\lambda$ with $\tau=2\operatorname{Re}\lambda$ and $\delta=|\lambda|^2>0$. Then both have negative real part iff $\tau<0$ (and $\delta>0$ holds automatically).
- *Real roots.* Both are negative iff their sum is negative and their product is positive: $\tau<0$ and $\delta>0$. (If the product is positive they have the same sign, and then the sum decides the sign.)

In both cases: Hurwitz $\iff\tau<0$ and $\delta>0$. Finally $\tau=2\operatorname{Re}z$ and $\delta=|z|^2-|w|^2$, so the condition reads $\operatorname{Re}z<0$ and $|w|<|z|$.

Part 2. Let $A\in\mathcal H$ with eigenvalues $\lambda_1,\lambda_2$ ($\operatorname{Re}\lambda_i<0$). For $t\in[0,1]$ let $M_t=(1-t)A-tI$. It depends continuously on $t$, $M_0=A$, $M_1=-I$. The eigenvalues of $M_t$ are $\mu_i(t)=(1-t)\lambda_i-t$ (if $Av=\lambda v$ then $M_tv=((1-t)\lambda-t)v$, and the characteristic polynomial of $M_t$ has these two roots with multiplicity). Their real parts are
$$\operatorname{Re}\mu_i(t)=(1-t)\operatorname{Re}\lambda_i-t<0\qquad(0\le t\le1),$$
since $(1-t)\operatorname{Re}\lambda_i\le0$ and $-t\le0$, and they cannot both vanish (at $t=0$ the first term is negative, for $t>0$ the second is). So $M_t\in\mathcal H$ for all $t$: every Hurwitz matrix is joined to $-I$ by a path in $\mathcal H$. Hence $\mathcal H$ is path connected.

Check. $A=\begin{pmatrix}-1&5\\0&-2\end{pmatrix}$: $\tau=-3<0$, $\delta=2>0$, eigenvalues $-1,-2$. At $t=\frac12$: $M_{1/2}=\begin{pmatrix}-1&2.5\\0&-1.5\end{pmatrix}$ with eigenvalues $-1,-1.5$, still Hurwitz.

Remark. In the coordinates, $\{|w|<|z|\}=GL_2^+$ and $\mathcal H$ is the part where $\operatorname{Re}z<0$. Writing $z=\rho e^{i\alpha}$ with $\alpha\in(\frac\pi2,\frac{3\pi}2)$, $\mathcal H\cong(\frac\pi2,\frac{3\pi}2)\times\{(\rho,w):\rho>|w|\}$, which is contractible. Note that the trace-determinant description is special to $n=2$; in higher dimensions Hurwitz stability needs the Routh-Hurwitz conditions.

Lesson. To prove that a set defined by the eigenvalues is path connected, find a homotopy that moves the eigenvalues in a controlled way (here an affine map $\lambda\mapsto(1-t)\lambda-t$).