Continuous images of connected spaces are connected (general Intermediate Value Theorem)

Step 1 — Suppose for contradiction that $f(X)$ is disconnected. Then $f(X) = A \cup B$ for some disjoint, nonempty sets $A, B$ that are each open in the subspace topology of $f(X)$: there exist open sets $A', B'$ in $Y$ with $A = f(X) \cap A'$ and $B = f(X) \cap B'$.

Step 2 — Pull the separation back to $X$. Let $P = f^{-1}(A')$ and $Q = f^{-1}(B')$. Since $f$ is continuous, $P$ and $Q$ are open in $X$. Every $x \in X$ has $f(x) \in f(X) = A \cup B$, so $f(x) \in A'$ or $f(x) \in B'$, meaning $x \in P$ or $x \in Q$: thus $X = P \cup Q$. Also $P$ and $Q$ are both nonempty (since $A, B$ are nonempty and are hit by $f$), and $P \cap Q = \emptyset$ (if $x \in P \cap Q$ then $f(x) \in A' \cap B' \cap f(X) = A \cap B = \emptyset$, impossible).

Step 3 — Contradiction. $P$ and $Q$ are disjoint nonempty open sets with $X = P \cup Q$, exactly the definition of $X$ being disconnected. This contradicts the hypothesis that $X$ is connected. So $f(X)$ cannot be disconnected: it is connected.

Step 4 — The interval corollary. The connected subsets of $\mathbb{R}$ are exactly the intervals (a standard fact: any subset that skips a real number between two of its points fails connectedness via the same open-set separation). Since $f(X)$ is connected by Steps 1–3, $f(X)$ is an interval of $\mathbb{R}$, so it contains every real number between any two of its elements $f(a)$ and $f(b)$ — the classical Intermediate Value Theorem, now seen as a special case of a purely topological fact.