Square-zero matrices form a path connected set

Idea. The set is a cone: stable under multiplication by scalars. Every point can be joined to the vertex $0$ by a straight segment inside the set.

Step 1: stability under scaling. For $A\in\mathcal N_2$ and $t\in\mathbb{R}$, $(tA)^2=t^2A^2=0$, so $tA\in\mathcal N_2$.

Step 2: path. For $A\in\mathcal N_2$ the map $\gamma(t)=(1-t)A$, $t\in[0,1]$, is continuous, with $\gamma(0)=A$, $\gamma(1)=0$, and stays in $\mathcal N_2$ by Step 1.

Step 3: any two points. Given $A,B\in\mathcal N_2$, join $A$ to $0$ and then $0$ to $B$ (reverse path). The concatenation is a path in $\mathcal N_2$ from $A$ to $B$. Hence $\mathcal N_2$ is path connected, indeed *star-shaped* with respect to $0$.

Not convex. For $n\ge2$ let $A=E_{12}$, $B=E_{21}$. Then $A^2=B^2=0$ but
$$\Bigl(\frac{A+B}2\Bigr)^2=\frac14(E_{12}+E_{21})^2=\frac14(E_{11}+E_{22})\ne0.$$
So the midpoint of $A$ and $B$ is not in $\mathcal N_2$ and $\mathcal N_2$ is not convex.

Remark. The same proof shows that $\{A:A^k=0\}$ (nilpotent of index $\le k$), the nilpotent cone $\{A:A^n=0\}$, and in fact every set of the form $\{A:f(A)=0\}$ with $f$ homogeneous (such as the singular matrices $\det=0$ or the rank $\le r$ matrices) is star-shaped with respect to $0$, hence connected. The set $\mathcal N_2$ is closed (polynomial equations) and unbounded: $\|sE_{12}\|=|s|$.

Lesson. A cone, i.e. a set stable under $A\mapsto tA$ for $t\in[0,1]$, is automatically contractible to its vertex, hence path connected.