Let $M_n(\mathbb R)$ denote the set of $n \times n$ real matrices with $n \ge 2$. Let $$\mathcal S = \{A \in M_n(\mathbb R): \rho(A) < 1\},$$ where $\rho(\cdot)$ denotes the spectral radius of a matrix. Then $\mathcal S$ is open and connected.
I am wondering whether $\mathcal S \setminus \{0\}$ is connected since the proof I know to show connectedness is every $A$ is connected to $0$.
As commented by Joe Johnson 126, if $n=1$ then this is clearly not true. I was in mind asking $n \ge 2$ so I edited the question to make it more clear.
Let $\lVert - \rVert'$ be a norm on $\mathbb{C}^n$ and $\lVert - \rVert$ be a compatible matrix norm on $M_n(\mathbb{C}) \supset M_n(\mathbb{R})$, i.e. $\lVert Ax \rVert' \le \lVert A \rVert \lVert x \rVert'$ for all $x \in \mathbb{C}^n$ and $A \in M_n(\mathbb{C})$. Then $\rho(A) \le \lVert A \rVert$.
Now let $A_1, A_2 \in \mathcal{S} \setminus \{ 0 \}$. We shall construct a path in $\mathcal{S} \setminus \{ 0 \}$ connecting $A_1$ and $A_2$.
We find continuous paths $u_i : [\varepsilon,1] \to \mathcal{S} \setminus \{ 0 \}, u_i(t) = t A_i$, connecting $B_i = \varepsilon A_i$ with $A_i$, such that $\lVert B_i \rVert \le 1$. The open set $C = \{ A \in M_n(\mathbb{R}) \mid \lVert A \rVert < 1 \}$ is contained in $ \mathcal{S}$. Hence it suffices to find a path in $C \setminus \{ 0 \}$ connecting $B_1$ and $B_2$. Since $C$ is convex, the line segment $s$ connecting $B_1$ and $B_2$ is contained in $C$. If $0 \notin s$, we are done. If $0 \in s$, we choose some $B \in C$ which is not contained in the line $l$ through $A_1, A_2$. This is possible because $n > 1$. Then the piecewise linear path going from $B_1$ to $B$ to $B_2$ is contained in $C \setminus \{ 0 \}$.