Let $n\ge 2$ and $V$ be an irreducible affine algebraic set in $\mathbb R^n$ . Then is it true that $V$ has only finitely many connected components in the Euclidean topology of $\mathbb R^n$ ? Does $\mathbb R^n \setminus V$ has only finitely many components in the Euclidean topology of $\mathbb R^n$ ?
[$V$ is actually a hypersurface since $V$ is the simultaneous zero set of finitely many polynomials, and we know that an $n$-tuple of real numbers is zero iff sum of their squares is zero]
Your suggestion about the sum of square is false, I mean it is not true that the dimension is $n-1$ even though your algebraic set is given by a single polynomial. For instance take the origin of $\mathbb{R}^2$ given as the zero locus of $x^2+y^2$. However, your question has a positive answer. I don't know if you are comfortable with semi-algebraic sets. What happens here is that an algebraic subset of $\mathbb{R}^n$ can be always retracted, by a deformation retraction, to an Euclidean closed and bounded semi-algebraic set. This happens by properties of continuous semi-algebraic mappings. Now every closed and bounded semi-algebraic set can be triangulated. Thus, you get that the homology of your starting real algebraic set is given by the simplicial homology of a finite simplicial complex. Therefore, every $H_k(V)$ has finite dimension, in particular $H_0(V)$. For the complement you get something similar because it is Zariski open in $\mathbb{R}^n$, thus, in the case of real algebraic sets, it is biregularly isomorphic to an algebraic set (it is exactly what happens for $\mathbb{R}\setminus\{0\}$ and the hyperbole). Then repeating the previous argument you get your result. Perhaps, being the complement, you can apply simpler set-theoretical arguments. Also in the case of algebraic sets there might be simpler proves. If you are interested in this topics I suggest you to look at chapter 9 of "Real Algebraic Geometry" by Boknach-Coste-Roy.