$[0,1]$ is compact without Heine-Borel

The bisection argument uses the nested-interval property of \(\RR\).

Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in A}\) be an open cover of \([0, 1]\). Suppose towards a contradiction that \([0, 1]\) cannot be covered by any finite subcollection of \(\mathcal{U}\). We say such an interval is uncoverable.

1. Bisection procedure:

   • Set \(I_0 = [a_0, b_0] = [0, 1]\), which is uncoverable.

   • Bisect \(I_0\) into two subintervals of equal length \(1/2\): \([0, 1/2]\) and \([1/2, 1]\). If both subintervals could be covered by finite subcollections \(\mathcal{U}_1\) and \(\mathcal{U}_2\) of \(\mathcal{U}\), then \(\mathcal{U}_1 \cup \mathcal{U}_2\) would be a finite cover of \([0, 1]\), a contradiction. Thus, at least one half is uncoverable. Let \(I_1 = [a_1, b_1]\) be an uncoverable half.

   • Inductively, having constructed an uncoverable interval \(I_k = [a_k, b_k]\) of length \(2^{-k}\), bisect it into \([a_k, \frac{a_k+b_k}{2}]\) and \([\frac{a_k+b_k}{2}, b_k]\). At least one of these two halves cannot be covered by any finite subcollection of \(\mathcal{U}\). Choose one such half and call it \(I_{k+1} = [a_{k+1}, b_{k+1}]\).

2. Intersection of nested intervals: This produces a sequence of nested closed intervals: \[ [0, 1] = I_0 \supset I_1 \supset I_2 \supset \cdots \supset I_k \supset \cdots \] with length \(\operatorname{len}(I_k) = 2^{-k} \to 0\) as \(k \to \infty\). By the Cantor Intersection Theorem (or least-upper-bound property of \(\RR\)), the intersection of these nested non-empty closed bounded intervals contains a unique point: \[ \bigcap_{k=0}^\infty I_k = \{x_0\} \quad \text{for some } x_0 \in [0, 1]. \]

3. Reaching a contradiction: Since \(\mathcal{U}\) covers \([0, 1]\), there is some open set \(U_{\alpha_0} \in \mathcal{U}\) containing \(x_0\). Since \(U_{\alpha_0}\) is open, there exists an open ball (interval) \(B(x_0, \varepsilon) = (x_0 - \varepsilon, x_0 + \varepsilon)\) such that: \[ B(x_0, \varepsilon) \cap [0, 1] \subseteq U_{\alpha_0}. \] Choose \(k \in \mathbb{N}\) large enough so that \(2^{-k} < \varepsilon\). Since \(x_0 \in I_k\) and \(\operatorname{len}(I_k) = 2^{-k} < \varepsilon\), for any point \(y \in I_k\): \[ |y - x_0| \leq \operatorname{len}(I_k) = 2^{-k} < \varepsilon. \] Thus \(I_k \subseteq B(x_0, \varepsilon) \cap [0, 1] \subseteq U_{\alpha_0}\).

   This means that \(I_k\) is covered by a single set \(U_{\alpha_0} \in \mathcal{U}\). This directly contradicts the fact that \(I_k\) is uncoverable by any finite subcollection of \(\mathcal{U}\).

Therefore, \([0, 1]\) must admit a finite subcover from \(\mathcal{U}\), proving that \([0, 1]\) is compact.