= Solution
Suppose first that $X$ is connected and $f:X\to\mathbb Z$ is continuous. Its image is connected by the <continuous image of a connected space>, but $\mathbb Z$ is discrete, so its only connected nonempty subsets are singletons. Hence $f$ is constant.
Conversely, if $X=U\cup V$ is a disconnection into nonempty disjoint <open sets>, the <function> equal to zero on $U$ and one on $V$ is continuous and nonconstant. This proves the <integer-valued function criterion for connectedness>.
Now let $f:X\to\mathbb Z$ be continuous under the hypotheses on the family $\mathcal A$. Each restriction $f|_A$ is constant because $A$ is connected. If $A,B\in\mathcal A$, a point of $A\cap B$ shows that their two constants agree. Since the sets cover $X$, $f$ is constant on $X$, and the criterion proves that $X$ is connected. This is the <pairwise-intersecting connected cover> argument.
Finally, fix $y_0\in Y$. For each $x\in X$, the set
$$
A_x=(X\times\{y_0\})\cup(\{x\}\times Y)
$$
is connected: its two connected pieces meet at $(x,y_0)$. The sets $A_x$ cover $X\times Y$ and any two share $X\times\{y_0\}$. The preceding result proves that $X\times Y$ is connected, giving the <product of connected spaces> result.
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