Regular elements form a connected dense open subset

By Regular elements and rank for a complex semisimple Lie algebra, $r$ is the minimum of the dimensions of the centralizers of elements of $\mathfrak{g}$. These dimensions belong to the finite set $\{0,\dots ,n\}$, and $\mathfrak{g}$ is nonempty, so the minimum is attained. Thus at least one regular element exists.

In a fixed basis, the matrix entries of ${\mathrm{ad}}_x$ depend linearly on $x$. Rank-nullity gives $\dim C_{\mathfrak{g}}(x)=n-\mathrm{rank}({\mathrm{ad}}_x)$. Hence regularity is the condition that the rank equals its attained maximum $n-r$. For $n-r>0$, the nonregular locus is the common zero set of the $(n-r)\times (n-r)$ minors. If $n-r=0$, every element is regular. In either case the regular locus is a nonempty Zariski-open set. A nonzero minor at a regular point remains nonzero on a Euclidean neighborhood; a polynomial vanishing on the regular locus therefore vanishes identically. Thus the regular locus is Zariski dense.

Let $x,y$ be regular. On the complex affine line through them, choose a maximal minor nonzero at $x$ (when $n-r=0$, the whole space is regular). Its restriction is a nonzero one-variable polynomial, so the nonregular points on that line form a finite set. A complex line with finitely many points removed is path connected, giving a path of regular elements from $x$ to $y$. The zero-dimensional case is a singleton. Thus the regular locus is connected.