 Solution

Let $C$ and $D$ be the connected components of $x$ and $y$. First, $C\times D$ is connected in the product topology. For each $c\in C$, the connected slice $\{c\}\times D$ meets the connected slice $C\times \{y\}$. Their union over $c$ is connected by the common-intersection argument, and is exactly $C\times D$.

Now let $K$ be the connected component of $(x,y)$ in $X\times Y$. The coordinate projections are continuous, so their images of $K$ are connected and contain $x$ and $y$. Maximality therefore gives $K\subseteq C\times D$. Conversely, the connected set $C\times D$ contains $(x,y)$, so it lies in $K$. $K=C\times D$. This is connected components of a product space; it requires neither local connectedness nor path connectedness.