證明計畫 · 由所求想起 所求是矛盾。反面假設交來兩條越靠越近、像卻分得開的數列；而 K bounded 讓 Bolzano-Weierstrass 從第一條裡挑出一條收斂的子數列，K closed 讓極限點留在 K 裡。第二條的對應項被距離拖著跑向同一個極限，於是兩邊的像都跑向同一個值——像的距離只好趨向零，與「始終不低於 ε₀」正面衝突。
Proof. Suppose f is not uniformly continuous on K. By 23.2 there are ε₀ > 0 and sequences (xₙ), (yₙ) in K with ‖xₙ − yₙ‖ < 1/n, ‖f(xₙ) − f(yₙ)‖ ≥ ε₀. Since K is bounded, so is (xₙ); by 16.4 some subsequence (x_{n(k)}) converges to a point z, and z ∈ K because K is closed.
這一步的所求：把反例數列逼出一個落腳點。兩個條件各管一件事——bounded 讓 §16-3 的 16.4（有界數列必有收斂子數列）有得用，closed 讓極限不會逃到 K 外面（10.5：closed 就是收齊所有 cluster point）。少了任何一個，接下來那句「f 在 z 連續」就沒有立足點，因為 z 可能根本不在定義域裡。
The corresponding terms of Y converge to the same point, since ‖y_{n(k)} − z‖ ≤ ‖y_{n(k)} − x_{n(k)}‖ + ‖x_{n(k)} − z‖, and both terms on the right tend to 0. By 20.2(c) both (f(x_{n(k)})) and (f(y_{n(k)})) converge to f(z).
Hence ‖f(x_{n(k)}) − f(y_{n(k)})‖ → 0, so for large k this quantity is smaller than ε₀ — contradicting the second relation above. Therefore f is uniformly continuous on K.
這一步結案。矛盾發生在一個看似最不起眼的地方：兩條數列的像各自跑向同一個值 f(z)，於是它們彼此的距離必定趨向零，而反例當初承諾這個距離永遠不低於 ε₀。要留意矛盾只需要「大的 k」一次，不必對所有 k 成立。
反例活不過 Bolzano-Weierstrass。∎
這張圖在說第一條路線的矛盾機制：反例交出的兩條數列成對出現、對應點越靠越近，而 K 的兩個條件合起來逼出一個共同的落腳點 z。連續於是讓兩邊的像同時跑向 f(z)，它們彼此的距離只能趨向零——這與反例承諾的「永遠隔著 ε₀」直接衝突。
Proof. Let ε > 0. For each u ∈ K, continuity at u gives a number d(u) > 0 such that ‖f(x) − f(u)‖ < ½ε whenever x ∈ K and ‖x − u‖ < d(u). Let G(u) be the open ball of radius ½d(u) about u.
The family {G(u) : u ∈ K} covers K, since each u lies in its own ball. By compactness finitely many of them cover K, say those centred at u₁, ⋯, u_N. Put δ(ε) = inf {½d(u₁), ⋯, ½d(u_N)}, which is positive because it is the smallest of finitely many positive numbers.
這一步是全證明的樞紐。無限多個正數的 inf 可以是零（§23-1 例 1 就是），有限多個正數的 inf 一定是正的——compact 買下的就是這個差別。要留意挑出來的是有限多個球，對應的 d(u) 也就只剩有限多個；沒有這一步，直接對所有 u ∈ K 取 inf 會落空。
Now let x, u ∈ K with ‖x − u‖ < δ(ε). Then x ∈ G(u_k) for some k, so ‖x − u_k‖ < ½d(u_k), and ‖u − u_k‖ ≤ ‖u − x‖ + ‖x − u_k‖ < d(u_k). Hence both ‖f(x) − f(u_k)‖ and ‖f(u) − f(u_k)‖ are less than ½ε, and adding gives ‖f(x) − f(u)‖ < ε.
Proof. Let ε > 0 and take δ(ε) > 0 from uniform continuity. Since (xₙ) is Cauchy there is M such that ‖xₙ − xₘ‖ < δ(ε) for all n, m ≥ M. For such n, m both points lie in A, so ‖f(xₙ) − f(xₘ)‖ < ε. Hence (f(xₙ)) is Cauchy.