Under the Zariski topology, this is true by basic Algebraic Geometry. However, I was wondering about how to prove this for the metric topology on $\mathbb{C}$. In *Algebraic Curves and Riemann Surfaces* by Miranda, this is stated as a fact, with a reference to Shafarevich (without giving page number). However, I can't quite see any obvious way to prove this.

More formally, we let $f(z,w)\in\mathbb{C}[z,w]$ be an irreducible polynomial, and we define $X=\{(z,w)\in\mathbb{C}^2\mid f(z,w)=0\}$, and give $X\subset\mathbb{C}^2$ the subspace topology. I wish to prove that given disjoint closed sets $X\_1$ and $X\_2$ s.t. $X=X\_1\cup X\_2$, either $X\_1=\emptyset$ or $X\_2=\emptyset$. I'm struggling to find the connection between the Zariski topology and the metric topology.