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The product of connected spaces is connected, so the product of the two connected components lies in the component of $(x,y)$. Conversely, continuous projections carry that component to connected sets in the factors, which must lie in $C_x$ and $C_y$. These two inclusions prove equality in arbitrary topological spaces.

True. For nonempty connected spaces $X,Y$, choose $x_0\in X$, $y_0\in Y$. For each $x\in X$, the subspace $(X\times \{y_0\})\cup (\{x\}\times Y)$ is connected: its two connected pieces meet at $(x,y_0)$. The union of these subspaces over $x$ is all of $X\times Y$, and every piece contains the common connected subset $X\times \{y_0\}$. Their union is therefore connected. The assertion is also true if a factor is empty, since the empty product space is connected. This proves product of connected spaces.

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 setis 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.