I am looking for a relatively "elementary" proof that every variety in ${\mathbb R}^n$ contains at least one non-singular point.

So far I only have such a proof for the case of hypersurfaces. Hopefully this proof would also explain what I mean by an "elementary" proof. Consider a variety $V \subset {\mathbb R}^n$ of dimension $n-1$, and note that the ideal $I(V)$ is generated by a single polynomial $f\in {\mathbb R}[x_1,\ldots,x_n]$. A point $p\in V$ is singular if the gradient $\nabla f$ vanishes on $p$. Without loss of generality, assume that $f$ depends on $x_1$, so $f_1 = \frac{\partial f}{\partial x_1}$ is not identically zero. If $f_1$ vanishes on $V$ then $f_1\in I(V)$, which is a contradiction to $I(V)$ being generated by $f$. Thus, there exists a point $p\in V$ such that $f_1(p)\neq 0$. By definition, $p$ is not a singular point of $V$.

I could not get a similar idea to work for lower-dimensional varieties. In case it helps, let me also add the definition of a singular point in this case. Consider a variety $V \subset {\mathbb R}^n$ of dimension $d$ and assume that $I(V)$ is generated by $f_1,\ldots,f_k$. Then a point $p\in V$ is singular if the rank of the Jacobian matrix of $f_1,\ldots,f_k$ at $p$ is smaller than $n-d$.

Real Algebraic Geometryby Bochnak, Coste, and Roy; also see Remark 3.3.15 for a warning about failure of density in the classical topology for this algebraic notion of smoothness. I suggest that you first read the Introduction to appreciate some of the subtleties of real algebraic geometry (since your argument for hypersurfaces is not quite convincing since it seems to be mixing up what is a definition and what is a theorem, so that it seems you might not be aware of the pitfalls for passing between geometry and algebra over $\mathbf{R}$). $\endgroup$3more comments