Strictly diagonally dominant matrices are invertible

Invertibility. Suppose $Ax=0$ with $x\ne0$. Choose $i$ with $|x_i|=\max_j|x_j|$; then $|x_i|>0$. The $i$-th equation reads $a_{ii}x_i=-\sum_{j\ne i}a_{ij}x_j$, so
$$|a_{ii}|\,|x_i|\le\sum_{j\ne i}|a_{ij}|\,|x_j|\le\Big(\sum_{j\ne i}|a_{ij}|\Big)|x_i|.$$
Dividing by $|x_i|>0$ gives $|a_{ii}|\le\sum_{j\ne i}|a_{ij}|$, contradicting dominance. So $\ker A=\{0\}$ and $A$ is invertible.

Openness. For each $i$, the function $f_i(A)=|a_{ii}|-\sum_{j\ne i}|a_{ij}|$ is continuous on $M_n(\mathbb{R})$ (a combination of absolute values of coordinates). Therefore
$$\mathcal{D}=\bigcap_{i=1}^n f_i^{-1}\big((0,\infty)\big)$$
is a finite intersection of open sets, hence open. By the first part $\mathcal{D}\subset GL_n(\mathbb{R})$.

Example. $\begin{pmatrix}3&1\\-1&2\end{pmatrix}$ is dominant ($3>1$, $2>1$) with determinant $7\ne0$.

Remark. This is the real core of Gershgorin's theorem: every eigenvalue lies in a disc $|z-a_{ii}|\le\sum_{j\ne i}|a_{ij}|$, since $A-zI$ would otherwise be strictly dominant. Dominance is only sufficient: $\begin{pmatrix}1&2\\0&1\end{pmatrix}$ is invertible but not dominant.