Proof of Intermediate Value Theorem on a Closed Real Interval

Case 1: $f(a)\le y\le f(b)$. Let

Since $f(a)\le y$ we have $a\in S$, so $S$ is nonempty; and $x\le b$ for every $x\in S$, so $S$ is bounded above by $b$. By the least upper bound property recorded in The Real Line: Standing Notation and Background for Calculus §completeness, $c=\sup S$ exists. Since $a\in S$ and $c$ is an upper bound for $S$, $a\le c$; since $b$ is an upper bound for $S$ and $c$ is the least one, $c\le b$. Hence $c\in[a,b]$, and $f$ is continuous at $c$, since continuity on $[a,b]$ means continuity at every point of $[a,b]$ by The Real Line: Standing Notation and Background for Calculus §continuity.

Step 1: $f(c)\le y$. Suppose instead that $y<f(c)$, and put $\varepsilon=f(c)-y$, so $\varepsilon>0$. By continuity of $f$ at $c$ there is $\delta>0$ such that every $x\in[a,b]$ with $|x-c|<\delta$ satisfies $|f(x)-f(c)|<\varepsilon$, and hence satisfies

Let $s\in S$. Then $s\le c$, because $c$ is an upper bound for $S$. If $c-\delta<s$, then $|s-c|=c-s<\delta$, so $f(s)>y$, contradicting $s\in S$; hence $s\le c-\delta$. Thus $c-\delta$ is an upper bound for $S$, and since $c$ is the least upper bound, $c\le c-\delta$, whence $\delta\le 0$ by Elementary Order Arithmetic in an Ordered Field. This contradicts $\delta>0$. Therefore $f(c)\le y$.

Step 2: $y\le f(c)$. Suppose instead that $f(c)<y$, and put $\varepsilon=y-f(c)$, so $\varepsilon>0$. Since $f(c)<y\le f(b)$ we have $f(c)\ne f(b)$, so $c\ne b$; combined with $c\le b$ this gives $c<b$. By continuity of $f$ at $c$ there is $\delta>0$ such that every $x\in[a,b]$ with $|x-c|<\delta$ satisfies $|f(x)-f(c)|<\varepsilon$, and hence satisfies

Let $x_0=\min\{b,\;c+\delta/2\}$. Since $c<b$ and $c<c+\delta/2$, we have $c<x_0$; and $x_0\le b$. Together with $a\le c<x_0$ this gives $x_0\in[a,b]$. Moreover $0<x_0-c\le\delta/2<\delta$, so $|x_0-c|<\delta$ and therefore $f(x_0)<y$. Hence $x_0\in S$, so $x_0\le\sup S=c$, contradicting $c<x_0$. Therefore $y\le f(c)$.

Steps 1 and 2 give $f(c)=y$, which proves Case 1.

Case 2: $f(b)\le y\le f(a)$. Define $g:[a,b]\to\mathbb{R}$ by $g(x)=(-1)\cdot f(x)$. By clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, $g$ is continuous on $[a,b]$. Multiplying the inequalities $f(b)\le y\le f(a)$ by $-1$ reverses them, by Elementary Order Arithmetic in an Ordered Field, so

Applying Case 1 to $g$ and to the value $-y$ yields $c\in[a,b]$ with $g(c)=-y$, that is, $-f(c)=-y$ and hence $f(c)=y$.