Assume without loss of generality that $f(a) < y < f(b)$. (If $f(a) > y > f(b)$, apply the result to $-f$.)

Construction of $S$: Define

$S = \{x \in [a, b] \mid f(x) < y\}.$

Step 1: $S$ is nonempty and bounded above.

• Since $f(a) < y$, we have $a \in S$, so $S \neq \varnothing$.
• Since $S \subseteq [a, b]$, $S$ is bounded above by $b$.

Step 2: By completeness, $c = \sup S$ exists, and $a \leq c \leq b$.

Step 3: We show $f(c) = y$ by eliminating the cases $f(c) < y$ and $f(c) > y$.

Case 1: $f(c) < y$ leads to contradiction.

Suppose $f(c) < y$. Let $\epsilon = y - f(c) > 0$. Since $f$ is continuous at $c$, there exists $\delta > 0$ such that

$|x - c| < \delta \quad \text{and} \quad x \in [a, b] \implies |f(x) - f(c)| < \epsilon/2.$

In particular, for $x \in (c, c + \delta) \cap [a, b]$,

$f(x) < f(c) + \epsilon/2 = f(c) + \frac{y - f(c)}{2} = \frac{f(c) + y}{2} < y.$

Thus $f(x) < y$ for all $x \in (c, c + \delta) \cap [a, b]$. This means that all points in $(c, c + \delta) \cap [a, b]$ belong to $S$.

Choose $x_0 = \min(c + \delta/2, b) > c$. Since $x_0 \in S$, we have $x_0 \leq \sup S = c$, contradiction. Thus $f(c) \not< y$.

Case 2: $f(c) > y$ leads to contradiction.

Suppose $f(c) > y$. Let $\epsilon = f(c) - y > 0$. By continuity at $c$, there exists $\delta > 0$ such that

$|x - c| < \delta \quad \text{and} \quad x \in [a, b] \implies |f(x) - f(c)| < \epsilon/2.$

For $x \in (c - \delta, c) \cap [a, b]$,

$f(x) > f(c) - \epsilon/2 = f(c) - \frac{f(c) - y}{2} = \frac{f(c) + y}{2} > y.$

Thus $f(x) > y$ for all $x \in (c - \delta, c)$. This means that no points in $(c - \delta, c)$ belong to $S$ (since $f(x) > y$ means $x \notin S$).

Therefore, $c - \delta$ is an upper bound for $S$ (any $x \in S$ with $x > c - \delta$ would satisfy $x \geq c$, contradicting $x \in (c - \delta, c)$). But $c - \delta < c = \sup S$, contradicting the definition of supremum. Thus $f(c) \not> y$.

Conclusion: Since $f(c) \not< y$ and $f(c) \not> y$, we must have $f(c) = y$.