It is sufficient for the proof to consider two connected spaces \(X\) and \(Y\). Suppose \(U\) and \(V\) disconnect \(X\times Y\) and assume \(U\) is non-empty, i.e. there is a point \((x_0,y_0)\in U\).

Since \(Y\) is connected and \(\{x_0\}\times Y\subseteq U\cup V\) this lemma implies that \(\{x_0\}\times Y\subseteq U\), i.e. \((x_0,y)\in U\) for every \(y\in Y\). With the same argument \(X\times \{y\}\subseteq U\) for every \(y\in Y\). Finally, \(\cup_{y\in Y} X\times \{y\}=X\times Y\subseteq U\). Therefore, \(X\times Y\) is connected.