= Solution
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 disconnection
$$
X=h^{-1}(\{0\})\cup h^{-1}(\{1\}).
$$
Conversely, 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 $\operatorname{Cl}(C)=U\cup V$ 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 $\operatorname{Cl}(C)$ is disjoint from $C$, contradicting $v\in\operatorname{Cl}(C)$. Therefore the <closure of a connected set> is connected.
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