This is a very nice problem I encountered in Analysis II as an undergraduate. It has a two-line solution using a theorem from Functional Analysis, but I didn’t know that at the time and managed to solve it via an elementary contradiction.
Problem: Show that the open unit interval \((0,1)\) is not the disjoint union of closed intervals of positive length.
Solution: Suppose that \[I = (0,1) = \bigsqcup_{J \in \mathcal{A}} J\] for some collection \(\mathcal{A}\) where all the \(J\) have positive length (in particular, they have distinct endpoints). Then first of all, this collection of intervals must be infinite, since the union of a finite number of closed sets is closed whereas \(I\) is not closed.
Next, let \(D_I\) be the set of all distances between pairs of intervals \(J, J'\). If this set had a lower bound \(\epsilon\) greater than zero, then \(|\mathcal{A}| \leq \lceil 1/\epsilon \rceil\) since at most that number of gaps fit inside \([0,1]\); this contradicts the fact that \(\mathcal{A}\) is infinite. Thus the infimum of \(D_I\) is \(0\); so we can find two distinct intervals \(L_1 = [a_1,l_1]\) and \(R_1 = [r_1,b_1]\) with \[a_1 < l_1 < r_1 < b_1 \textrm{ and } r_1 - l_1 < 1/2.\]
Let \(I_1 = (l_1, r_1)\) be the open interval between \(L_1\) and \(R_1\), and \(\mathcal{A}_1\) the subset of \(\mathcal{A}\) of intervals that intersect \(I_1\). Then these cover \(I_1\), are disjoint and do not intersect \(L_1\) or \(R_1\), so \[I_1 = \bigsqcup_{J \in \mathcal{A}_1}J.\] As before, \(\mathcal{A}_1\) is infinite, so if \(D_1\) is the set of distances of members of \(\mathcal{A}_1\), it too has zero infimum. Therefore, we can repeat the previous construction to find intervals \(L_2 = [a_2,l_2]\) and \(R_2 = [r_2,b_2]\) bordering the open interval \(I_2 = (l_2,r_2)\) such that \(|I_2| = r_2 - l_2 < 1/4\).
Continuing this process we find \(L_k, R_k \in \mathcal{A}\) and construct a nested sequence of intervals \(I_k = (l_k, r_k)\) of length \(|I_k| < 2^{-k}\). The shrinking lengths of the \(I_k\) imply that the sequence \[s := l_1,\, r_1,\, l_2,\, r_2,\,\ldots\] is Cauchy; therefore it has a limit \(x \in (0,1).\) Furthermore, the sub-sequence \(l_k\) is strictly increasing, so \(l_k \nearrow x\), and the sub-sequence \(r_k\) is strictly decreasing, so \(r_k \searrow x\).
Since \(\mathcal{A}\) partitions \((0,1)\), there exists a unique \(J = [l,r] \in \mathcal{A}\) such that \(x \in J\). Let \(\epsilon\) be the length of \(J\): \(\epsilon=r -l\). Since the sequence \(s\) converges to \(x\), after a certain \(k_0\) we have \(0 < x - l_k < \epsilon/2\) and \(0 < r_k - x < \epsilon/2\). Depending on which half of \(J\) contains \(x\), either the first or the second (or both) inequality gives a contradiction: infinitely many \(L_k\) or \(R_k\) intersect \(J\), whereas \(J\) can be at most one of those.