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Suppose for contradiction that the continuous image of a connected space $Y=f(X)$ is disconnected. Then there are disjoint nonempty sets $U,V$, open in the subspace topology on $Y$, with $Y=U\cup V$. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open subsets of $X$; by surjectivity both are nonempty, and their union is $X$. This is a separation of a topological space of $X$, contradicting that $X$ is a connected space. Hence $Y$ is connected.

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.

A subset $A\subseteq X$ is connected if it cannot be written as $A=U\cup V$, where $U,V$ are disjoint, nonempty sets open in the subspace topology on $A$.

If $f(X)$ were disconnected as $U\cup V$, continuity would make $f^{-1}(U)$ and $f^{-1}(V)$ a disconnection of $X$. Thus the continuous image of a connected space is connected.

For $Y=\{0,1\}$ with the discrete topology, a nonconstant continuous $h:X\to Y$ gives the disconnectionConversely, a disconnection $X=U\cup V$ defines a continuous nonconstant function by assigning $0$ on $U$ and $1$ on $V$. Hence $X$ is connected exactly when every such $h$ is constant.

Finally suppose $C$ is connected but $\mathrm{Cl}(C)=U\cup V$ is a disconnection. The intersections $C\cap U$ and $C\cap V$ cannot both be nonempty, so, after interchanging $U,V$, we have $C\subseteq U$. For any $v\in V$, the relatively open neighbourhood $V$ of $v$ in $\mathrm{Cl}(C)$ is disjoint from $C$, contradicting $v\in \mathrm{Cl}(C)$. Therefore the closure of a connected set is connected.