Any Real Algebraic Variety Has Finitely Many Path Components? Consider the following statement:
Statement: Any real algebraic variety has only finitely many path components.
By 'path component' I mean: Let $X$ be a real algebraic variety. Then $X$ is a subset of some Euclidean Space, say $\mathbb R^n$. Then the path components of $X$ under the usual topology of $\mathbb R^n$ are the 'path components of the real algebraic variety $X$'.
It is known that any algebraic variety is a union of finitely many irreducible algebraic varieties.
So we need to focus only on irreducible algebraic varieties.
The set $S=\{(x,y)\in\mathbb R^2:x^4+y^4=x^2+y^2\}$ is an irreducible algebraic variety having two path components.
So it is not true that irreducible real algebraic varieties are path connected.
But it might still be true that irreducible real algebraic varieties have only finitely many path components, which would lead to the conclusion that the 'statement' given in the beginning is true.
Does anybody know the answer to this?
(I have an extremely limited knowledge of algebraic geometry so please forgive me if my question is a dumb one.)
Thanks.

EDIT:
The particular type of algebraic variety I am interested in is defined below:
Consider the space $(\mathbf R^n)^m=\overbrace{\mathbf R^n\times\cdots\times\mathbf R^n}^{m\text{ times}}$.
Let $E$ be a subset of $\{\{i,j\}\in\mathbb N^2:1\leq i<j\leq m\}$.
For each $\{i,j\}\in E$, let $l_{ij}$ be a fixed positive real number.
For each $\{i,j\}\in E$ define a function $f_{ij}:(\mathbf R^n)^m\to \mathbf R$ as $f_{ij}(\mathbf x)=|\mathbf x_i-\mathbf x_j|^2-l_{ij}^2$.
(Here $\mathbf x=(\mathbf x_1,\ldots,\mathbf x_m)\in(\mathbf R^n)^m$ and $|\mathbf x_i-\mathbf x_j|$ is the Euclidean distance between the points $\mathbf x_i$ and $\mathbf x_j$.) 
Define $X=\bigcap_{\{i,j\}\in E}f_{ij}^{-1}(0)$.
Then $X$ is a real algebraic variety.
What can we say about the number of path components of $X$?
 A: Yes it is known, this is theorem 4.1 of this article by Delfs and Knebusch (I dont know if you can access it though, i can through my lab)
I've checked out the proof it does use a fair deal of algebraic geometry.
The proof goes like this.
First you reduce to the case of integral affine varieties (which seems to be the interesting case for you).
You then have a finite surjective morphism from $X$ to $\mathbb A^n$, this is Noether's normalisation lemma. The $n$ is also the dimension of $X$, and you now proceed by induction on $n$.
Now the key point is the following fact
Lemma: If $f:X\to Y$ is a finite morphism between two real algebraic variety and if $B$ is a path component of $X$, such that $f: B\to Y$ is étale, then the image of $B$ is a connected component of $Y$, and the fibers of $B\to f(B)$ all have the same number of (real) points.
This tells you that outside the ramification locus of the map given by Noether's lemma then $X$ has exactly $d$ real points in each fiber. If this ramification locus is empty then you're done by the lemma, because the paths connected components should all map to $\mathbb R^n$ and you have thus at most $d$ of them.
If the ramification locus is not empty, then the branch locus (the inverse image of the ramification locus) is of dimension n-1 and can thus be assumed to have finitely many path components.
Now if you take $f(x)$ a point outside of the ramification locus, and a line passing through this point and any point of the ramification locus, the intersection will be a finite number of points, thus you can find and interval on the line of the form $[f(x), y]$ with f ramified only over some point of the preimage of $y$.
The other key point is that, in such conditions, such a path can be lifted to a path on $X(\mathbb R)$ (it is a technical result in the same spirit as the lemma afore mentionned), and thus you can connect every point outside the branch locus to a point in the branch locus. Thus $X$ has at most as many path-component as the branch locus, which has only a finite number of them by induction.
I know this is not elementary but i doubt you can find a more elementary treatment (althgouh i am in no mean an expert on this subject).
