Precursor definitions and proposition, from Bredon's Topology and Geometry.
Definition: Let $X$ be a topological space. A separation of $X$ is a pair $U,V$ of disjoint, nonempty, open subsets of $X$ whose union is $X$. We say that $X$ is disconnected if such a separation exists, and say it is connected otherwise.
Definition: Let $X$ be a space. A discrete valued map is a continuous function $f:X \rightarrow D$ where $D$ is a discrete space, i.e a topological space with the discrete topology.
Proposition: A space $X$ is connected if and only if every discrete valued map $f:X \rightarrow D$ is constant.
Proof: It suffices to show that $X$ is disconnected if and only if there exists a discrete map $f:X \rightarrow D$ which is not constant. Suppose $A,B \subseteq X$ constitute a separation of $X$. Define a map $f:X \rightarrow \{0,1\}_{\text{disc}}$ by
$$
f(x) =
\begin{cases}
1 & \text{ if } x \in A \\
0 & \text{ if } x \in B
\end{cases}
$$
Then $f$ is continuous; one readily checks $f^{-1}[\varnothing] = \varnothing, f^{-1}[\{0,1\}] = X, f^{-1}[0] = B, f^{-1}[1] = A$, all of which are open in $X$. It is clear that $f$ is not constant, take any $a \in A$ and $b \in B$, which are guaranteed to be distinct since $A \cap B = \varnothing$, then $f(a) = 1$ while $f(b) = 0$. Therefore $f$ is a discrete valued map which is non-constant. Suppose that $f:X \rightarrow Y$ is a discrete valued map which is not constant, and let $x,y\in X$ be two distinct points of $X$ for which $f(x) \neq f(y) \in D$. Since $D$ is a discrete space, singletons are open so that $f^{-1}[f(x)]$ is open in $X$. Remark that $D-\{f(x)\}$ is a subset of $D$, and as such is open in $D$, therefore $f^{-1}[D-\{f(x)\}] = X-f^{-1}[f(x)]$ is open in $X$. This means that $f^{-1}[f(x)]$ is a clopen set of $X$. It is clearly not-empty, and moreover since $y$ is a point distinct from $x$ not mapping to $f(x)$, it is also not all of $X$. Therefore $X$ is disconnected.
I can see that the proof of this statement works, but I don't see morally why one would ever come up with this connection in the first place. What is the intuitive reason that a connected space forces discrete maps out of it to be constant, beyond the fact that if one wasn't you could create a separation of $X$? Please let me know if I can be more clear, and thank you in advance.
Best Answer
If $X$ is connected and $f$ on $X$ is continuous, then the image of $f$ is connected. A connected discrete space cannot have more than one point.