Theorem 1.12 (Intermediate Value Theorem). Let $a<b$ be real numbers, and let $f : [a, b] →ℝ$ be continuous. Then $f$, in the interval $[a, b]$, takes on all values between $f (a)$ and $f (b)$.
Proof. Let $f(a)=α$ and $f(b)=β$, and let $γ$ be between $α$ and $β$. We may assume that $γ$ is in fact strictly between $α$ and $β$. Possibly by replacing $f$ by $−f$, we may assume that also $α<β$. Let $$A=\{x∈[a,b]| f(x)≤γ\}$$ and $$B=\{x∈[a,b]| f(x)≥γ\}.$$ These sets are nonempty, since $a$ is in $A$ and $b$ is in $B$, and $f$ is bounded as a result of Theorem 1.11. Thus the numbers $γ_1 = \sup\{f(x) | x ∈ A\}$ and $γ_2 = \inf \{ f (x) | x ∈ B\}$ are well defined and have $γ_1 ≤γ≤γ_2$.
If $γ_1 = γ$, then we can find a sequence $\{x_n\}$ in $A$ such that $f(x_n)$ converges to $γ$. Using the Bolzano-Weierstrass Theorem, we can find a convergent subsequence $\{x_{n_k}\}$ of $\{x_n\}$, say with limit $x_0$. By continuity of $f$, $\{f(x_{n_k})\}$ converges to $f(x_0)$. Then $f(x_0) = γ_1 = γ$, and we are done. Arguing by contradiction, we may therefore assume that $γ_1 < γ$. Similarly we may assume that $γ< γ_2$, but we do not need to do so.
Let $ε = γ_2 −γ_1$, and choose, by Theorem 1.10 and uniform continuity, $δ> 0$ such that $|x_1 − x_2| < δ$ implies $| f (x_1) − f (x_2)| < ε$ whenever $x_1$ and $x_2$ both lie in $[a, b]$. Then choose an integer $n$ such that $2^{−n} (b − a) < δ$, and consider the value of $f$ at the points $p_k = a+k2^{−n}(b−a)$ for $0 ≤ k ≤ 2^n$. Since $p_{k+1}−p_k = 2^{−n}(b−a)<δ$, we have $|f(p_{k+1})−f(p_k)|<ε= γ_2−γ_1$. Consequently if $f(p_k)≤γ_1$, then $$f(p_{k+1})≤ f(p_k)+|f(p_{k+1})− f(p_k)|< γ_1 +(γ_2 −γ_1)= γ_2,$$and hence $f (p_{k+1}) ≤γ_1$. Now $f (p_0) = f (a) = α≤γ_1$. Thus induction shows that $f(p_k)≤γ_1$ for all $k≤2^n$. However, for $k=2^n$, we have $p_{2^n} =b$, and $f (b) = β≥γ> γ_1$, and we have arrived at a contradiction.
This is a proof of the Intermediate Value Theorem given in Anthony W. Knapp’s Basic Real Analysis and repeated in his Basic Algebra. I’m good with the overall structure of the proof and the details of the first part, but I’m having trouble with the second part, the proof by contradiction. In particular, I can’t seem to figure out how he arrived at $f (p_{k+1}) ≤γ_1$ from the chain of inequalities immediately before. I’m probably missing something really obvious. Would appreciate any help!