Baire Category Theorem

Assume first that $X \ne \varnothing$ — if $X = \varnothing$ then $X = \bigcup_n \varnothing$, so the empty space is meagre in itself and nonemptiness is essential to the theorem.

Since $E_1$ is nowhere dense, $\operatorname{int}(\overline{E_1}) = \varnothing$, so $U_1 = X \setminus \overline{E_1}$ is a nonempty open set (if $U_1 = \varnothing$ then $\overline{E_1} = X$, giving $\operatorname{int}(\overline{E_1}) = X \ne \varnothing$).

Pick $x_1 \in U_1$; some $\rho > 0$ has $B(x_1, \rho) \subseteq U_1$, and with $r_1 = \min(\rho/2, 1/2)$ the closed ball $\overline{B}(x_1, r_1) \subseteq U_1$ with $r_1 < 1$.

Inductively, suppose $\overline{B}(x_k, r_k)$ is chosen.

Then $V = B(x_k, r_k) \setminus \overline{E_{k+1}}$ is a nonempty open set: $B(x_k, r_k)$ is nonempty open, and $B(x_k, r_k) \subseteq \overline{E_{k+1}}$ would give $B(x_k, r_k) \subseteq \operatorname{int}(\overline{E_{k+1}}) = \varnothing$, a contradiction.

Pick $x_{k+1} \in V$ and shrink the radius below $r_k/2$ so the closed ball fits in $V$: $\overline{B}(x_{k+1}, r_{k+1}) \subseteq B(x_k, r_k) \setminus \overline{E_{k+1}}$ with $0 < r_{k+1} < r_k/2$.

The sequence $(x_k)$ is Cauchy: for $m > k$, $x_m \in \overline{B}(x_{k+1}, r_{k+1}) \subseteq B(x_k, r_k)$, so $d(x_k, x_m) < r_k$, and $r_k \le 2^{-(k-1)} r_1 \to 0$ (equality at $k = 1$; for $k \ge 2$ iterating $r_{j+1} < r_j/2$ gives the strict inequality). Completeness of $X$ gives a limit $x_k \to x \in X$.

For fixed $k$, the tail $(x_m)_{m \ge k}$ lies in the closed set $\overline{B}(x_k, r_k)$, so the limit $x \in \overline{B}(x_k, r_k)$. Hence $x \in \bigcap_k \overline{B}(x_k, r_k)$ for all $k$; with $x \in \overline{B}(x_1, r_1) \subseteq X \setminus \overline{E_1}$ and, for $j \ge 2$, $x \in \overline{B}(x_j, r_j) \subseteq B(x_{j-1}, r_{j-1}) \setminus \overline{E_j}$, we get $x \notin E_j$ for every $j$.

But $x \in X = \bigcup_j E_j$, so $x \in E_j$ for some $j$ — a contradiction.

Hence a complete metric space cannot be a countable union of nowhere dense sets.

The intuition: each nowhere dense set leaves a "hole", and completeness traps the limit point obtained by descending through those holes inside the space.

By construction each $I_k$ is a closed interval with $I_{k+1} \subseteq I_k$; when $q_k$ lands in $I_{k-1}$ we take $I_k$ of one-quarter the length with $q_k \notin I_k$ (if $q_k$ already lies outside, $I_k = I_{k-1}$).

The lengths tend to $0$.

Suppose no more "hits" occur after stage $K$; then $I_n = I_K$ for all $n \ge K$ and the length stays positive.

But $I_K$ has positive length, so it contains infinitely many rationals, each appearing exactly once in the enumeration — hence some $q_m \in I_K$ has index $m > K$.

Then $I_{m-1} = I_K \ni q_m$, so stage $m$ is a hit — a contradiction.

Hits therefore occur infinitely often, and each one quarters the length, so $\ell_k \to 0$.

By completeness of $\mathbb{R}$ (the nested interval theorem), nested closed intervals whose lengths tend to $0$ intersect in exactly one point: $\bigcap_k I_k = \{x\}$. And since $x \in I_k$ while $q_k \notin I_k$, we have $x \ne q_k$ for every $k$.

Since $\{q_k\}$ enumerates every rational of $[0, 1]$, $x$ is an irrational in $[0, 1]$.

Hence $[0, 1] = \bigcup_k \{q_k\}$ — a countable union of points — is impossible, and $[0, 1]$ is uncountable.

This is the point-set version of Baire's theorem: a singleton is nowhere dense in $\mathbb{R}$, so $[0, 1]$, being complete, cannot be a countable union of points without contradicting Baire.

[Sketch] $F_m$ is closed. If $f_j \in F_m$ with $f_j \to f$ uniformly, witnessed by $x_j$ — i.e. $|f_j(x_j) - f_j(y)| \le m|x_j - y|$ — pass to a convergent subsequence $x_{j'} \to x$ (Bolzano–Weierstrass); the estimate survives the limit as $|f(x) - f(y)| \le m|x - y|$, so $f \in F_m$.

[Sketch] $F_m$ is nowhere dense.

Given $f \in F_m$ and $\varepsilon > 0$, first approximate $f$ within $\varepsilon/2$ by a piecewise-linear $p$ (uniform continuity), then superimpose a fine zigzag of amplitude $<\varepsilon/2$ whose slopes exceed those of $p$ by more than $m$.

At every candidate point $x$, a nearby tooth of the zigzag breaks the $m$-Lipschitz bound, so the perturbed $g \notin F_m$ while $\|g - f\|_\infty < \varepsilon$.

Being closed, $F_m$ has empty interior.

A function differentiable at even one point lies in $\bigcup_m F_m$. If $f'(x_0)$ exists, then $|f(y) - f(x_0)| \le m_1|y - x_0|$ for $|y - x_0| < \delta$ (take $m_1 \ge |f'(x_0)| + 1$), while for far points $|y - x_0| \ge \delta$ boundedness gives $|f(y) - f(x_0)| \le 2\|f\|_\infty \le \frac{2\|f\|_\infty}{\delta}|y - x_0|$. With an integer $m \ge \max(m_1, 2\|f\|_\infty/\delta)$ we get $|f(y) - f(x_0)| \le m|y - x_0|$ for all $y$ — i.e. $f \in F_m$.

Hence "functions differentiable somewhere" $\subseteq \bigcup_{m=1}^{\infty} F_m$ is a countable union of nowhere dense sets — meagre.

Its complement, the nowhere-differentiable functions, is comeager, and by Baire (problem 1) it is not just nonempty but dense.

To stress again: the closed/nowhere-dense verifications above are idea-level sketches, not complete proofs.