Let $X$ be an affine algebraic set, i.e. $X=V(I)$ is the zero set of some ideal $I\subseteq k[T_1,\ldots T_k]$. $X$ must not necessarily be irreducible.
Is there a finite decomposition into irreducible algebraic sets, i.e. $$ X = U_1\sqcup \ldots \sqcup U_n?$$
Summarizing @KReisers and my comments:
When $X=V(I)$ is an algebraic set, i.e. the zero locus of an ideal $I\subseteq k[T_1,\ldots T_k]$, it can be expressed as the finite union of irreducible algebraic varieties $$ X=\bigcup U_i$$ where $U_i=V(P_i)$ are the zero loci of prime ideals $P_i\subset k[T_1,\ldots T_k]$. This is a consequence of the Lasker-Noether primary decomposition.
However, this union need not be disjoint. Consider $X=V(xy)$ for $(xy)\subseteq k[x,y]$. Then $V(xy)=V(x)\cup V(y)$, where both $V(x)$ and $V(y)$ are irreducible. However, they are not disjoint, as $(0,0)$ is a common point.