How to prove that $ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $

Let $A\in M_n$. Choose $\epsilon >0$ such that $\epsilon$ is less than the smallest positive eigenvalue of $A$ (if any exists, otherwise any positive $\epsilon$ will do). Observe that $A-\epsilon I$ is invertible (otherwise, $\epsilon$ is an eigenvalue for $A$), therefore $A_\epsilon \in GL_n$.

I particularly like this argument. Take any matrix $A$ and any invertible matrix $B$. Consider $$p(t) = \det((1-t)A+tB)$$ which is clearly a polynomial in one variable, and it is not zero polynomial because $p(1)\neq 0$. Thus, it has finitely many zeroes. That means that you can find arbitrarily small $t$ such that $C_t = (1-t)A+tB$ is invertible and we have $$\|A-C_t\| = |t|\|A-B\|<\varepsilon,\quad |t|<\frac{\varepsilon}{\|A-B\|}$$ Geometrically, we consider a line segment through matrices $A$ and $B$ and can find invertible matrix on it arbitrarily close to $A$.

The function $\det g$ is a polynomial in the $g_{ij}$. In particular, it's equal to its Taylor series everywhere. It follows that the closed set $X = \{g\in M_n(\mathbb{R}):\, \det g = 0\}$ has empty interior, and thus its complement in $GL_n(\mathbb{R})$ is dense in $\mathbb{R}^n$.