**Proof Plan.** The proof proceeds in two steps: first show that $f([a,b])$ is connected (as the [continuous](/page/Continuity) image of the connected set $[a,b]$), then use the fact that connected subsets of $\mathbb{R}$ are intervals. **Step 1 (The image is connected).** The interval $[a,b]$ is connected (it is an interval in $\mathbb{R}$). The [continuous image of a connected space is connected](/theorems/296), so $f([a,b])$ is a connected subset of $\mathbb{R}$. **Step 2 (Connected subsets of $\mathbb{R}$ are intervals).** A subset $S \subseteq \mathbb{R}$ is connected if and only if it is an interval: whenever $s_1, s_2 \in S$ with $s_1 < s_2$, every $t$ with $s_1 < t < s_2$ satisfies $t \in S$. (If some $t \in (s_1, s_2)$ were missing from $S$, then $S = (S \cap (-\infty, t)) \cup (S \cap (t, \infty))$ would be a disconnection.) **Step 3 (Conclude).** Since $f([a,b])$ is a connected subset of $\mathbb{R}$, it is an interval. Both $f(a)$ and $f(b)$ belong to $f([a,b])$, so every value between $f(a)$ and $f(b)$ belongs to $f([a,b])$. In particular, for every $\gamma$ between $f(a)$ and $f(b)$, there exists $c \in [a,b]$ with $f(c) = \gamma$. $\blacksquare$