Let $X$ be a path connected topological space. Suppose that $X = A \cup B$ where $A$ and $B$ are non empty, disjoint, open sets. Let $a \in A$ $b \in B$ and let $\gamma: I \rightarrow X$ denote a path from $a$ to $b$

We have $I = \gamma^{-1}(A) \cup \gamma^{-1}(B)$ where $\gamma^{-1}(A),\gamma^{-1}(B)$ are non empty, open and disjoint. Since $I$ is connected, this is a contradiction, which concludes the proof.