[proofplan] We express $X \times Y$ as a union of connected sets that pairwise intersect. For each $y \in Y$, define $S_y$ as the union of the "horizontal base slice" $X \times \{b\}$, the "vertical connector" $\{a\} \times Y$, and the "horizontal target slice" $X \times \{y\}$, where $(a,b)$ is a fixed base point. Each $S_y$ is connected (it is a union of three connected sets sharing pairwise intersections), and all $S_y$ contain $(a,b)$, so they pairwise intersect. The [union of overlapping connected sets](/theorems/298) is connected. [/proofplan] [step:Construct connected subsets $S_y$ that cover $X \times Y$ and share the common point $(a,b)$] Fix $(a, b) \in X \times Y$. For each $y \in Y$, define \begin{align*} S_y = (X \times \{b\}) \cup (\{a\} \times Y) \cup (X \times \{y\}). \end{align*} Each of the three pieces is connected: $X \times \{b\}$ is homeomorphic to $X$ via $(x, b) \mapsto x$, so it is connected; $\{a\} \times Y$ is homeomorphic to $Y$ via $(a, y) \mapsto y$, so it is connected; and $X \times \{y\}$ is homeomorphic to $X$, so it is connected. The three pieces have pairwise non-empty intersections: $X \times \{b\}$ and $\{a\} \times Y$ share $(a, b)$; $\{a\} \times Y$ and $X \times \{y\}$ share $(a, y)$. By the [union of overlapping connected sets](/theorems/298) result, $S_y$ is connected. [/step] [step:Apply the union of overlapping connected sets to conclude $X \times Y$ is connected] Every point $(x, y) \in X \times Y$ belongs to $X \times \{y\} \subseteq S_y$, so \begin{align*} X \times Y = \bigcup_{y \in Y} S_y. \end{align*} Every $S_y$ contains the point $(a, b)$ (since $(a, b) \in X \times \{b\} \subseteq S_y$), so any two sets $S_y$ and $S_{y'}$ satisfy $S_y \cap S_{y'} \neq \varnothing$. By the [union of overlapping connected sets](/theorems/298) result, $X \times Y$ is connected. [guided] The direct approach --- trying to show that $X \times Y$ has no disconnection --- is awkward because the product topology mixes the two factors. Instead, we decompose $X \times Y$ into pieces whose connectedness reduces to the connectedness of $X$ and $Y$ individually. The idea is to "wire" every horizontal slice $X \times \{y\}$ to a base slice $X \times \{b\}$ through a vertical connector $\{a\} \times Y$. The set $S_y = (X \times \{b\}) \cup (\{a\} \times Y) \cup (X \times \{y\})$ looks like a letter "H" rotated $90$ degrees: two horizontal bars connected by a vertical bar. **Connectedness of each $S_y$:** Each horizontal bar $X \times \{y'\}$ is homeomorphic to $X$ via $(x, y') \mapsto x$, hence connected. The vertical bar $\{a\} \times Y$ is homeomorphic to $Y$ via $(a, y') \mapsto y'$, hence connected. The three pieces share points pairwise: $X \times \{b\}$ and $\{a\} \times Y$ share $(a, b)$; $\{a\} \times Y$ and $X \times \{y\}$ share $(a, y)$. By the [union of overlapping connected sets](/theorems/298), $S_y$ is connected. **Covering $X \times Y$:** Every point $(x, y) \in X \times Y$ belongs to $X \times \{y\} \subseteq S_y$, so $X \times Y = \bigcup_{y \in Y} S_y$. **Pairwise intersection:** Every $S_y$ contains the base point $(a, b)$ (since $(a, b) \in X \times \{b\} \subseteq S_y$), so any two sets $S_y$ and $S_{y'}$ satisfy $S_y \cap S_{y'} \supseteq \{(a, b)\} \neq \varnothing$. By the [union of overlapping connected sets](/theorems/298), $X \times Y$ is connected. [/guided] [/step]