Prove that $[0,1]$ is not a disjoint union of a countable family of closed sets [closed]

Question

This proof is probably too long, please don't downvote this. I'm just leaving it here in case somebody wants to actually verify its validity for me. I didn't use a hint in the book and decided to prove it my own way.

Outline of the proof:

• After trying to cover $[0,1]$ with 2 disjoint closed sets A and B, show that there will remain some open interval $(c_{0}, c_{1}) \in [0,1]$ that has to be covered, where $c_{0} \in A$ or $B$, $c_{1} \in A$ or $B$.
• Use some closed set to try cover this interval and obtain 2 intervals $(c_{00}, c_{01}), (c_{10}, c_{11})$ yet to cover, where all endpoints belong to closed sets (endpoints are not necessarily infs/sups of those closed sets, just points belonging to them).
• Repeat this process infinitely, proving that trying to cover each of $n$ intervals will create at least $2n$ yet uncovered open intervals (taking into account that total length of all open intervals has to approach 0).
• Prove that there will be an uncountable number of points left not belonging to any of the above closed sets after the above process, and name this set $N$. Name set of all endpoints of closed sets (i.e. $c_{0}, c_{10}$...) $E$.
• To make proving that $N$ is not a union of a countable number of closed sets easier, define a metric space $S$ with Euclidian metric which consists of a) set $N^\prime$ which contains same points as $N \in \mathbb{R}$, b) set $E^\prime$ which contains same points as $E \in \mathbb{R}$. Proceed to prove this space is complete, then assume $N^\prime$ is a union of a countable number of closed sets, and then use Baire Category Theorem to derive a contradiction, henceforth negating the assumption.
• Prove that $N^\prime$ not being a union of a countable number of closed sets in $S$ implies that $N$ is not a union of a countable number of closed sets in $\mathbb{R}$, which would imply that $[0,1]$ is not a union of a countable number of disjoint closed sets.

Proof:

1. Every nonempty bounded subset of $\mathbb{R}$ has inf/sup, and that inf/sup belongs to the subset if it is closed (if it's isolated from the set then it is a min/max of the set and is an isolated point of the set, if it's not isolated from the set then it's a limit point thus belongs to it).
2. Take a closed set $A \subset \mathbb{R} \mid A \cap [0,1] \neq \emptyset$. Take a random point $x \in ([0,1] \setminus A), x \notin \{0,1\} \implies \exists \epsilon > 0 \mid B_{\epsilon}(x) \in ([0,1] \setminus A)$, since if not, A will not be a closed set. Take $(c_{0}, c_{1}) \subset [0,1] \mid c_{0} = \max(\sup(A \cap [0,x)),0), c_{1} = \min(\inf(A \cap (x,1]), 1)$. If $c_{0} \notin A$ or $c_{1} \notin A$, then $c_{0} = 0$ or $c_{1} = 1$, so attempting to cover $[0, c_{1})$ or $(c_{0}, 1]$ with another closed set B will yield an open interval of form $(c_{0}, c_{1})$. Next:
3. Take arbitrary closed set $C_{0} \cap (c_{0},c_{1}) \neq \emptyset $. $c_{0},c_{1}$ are not limit points of $C_{0}$ as $c_{0},c_{1} \notin C_{0}$ since they belong to $B$ and $A$, also $C_{0} \cap (c_{0},c_{1})$ is bounded. Thus $\exists c_{01}, c_{10} \in C_{0} | c_{0} < \inf(C_{0} \cap (c_{0},c_{1}))=c_{01} \in C_{0} \leq \sup(C_{0} \cap (c_{0},c_{1}))=c_{10} \in C_{1} < c_{1}$. Denote $c_{0}$ as $c_{00}$ and $c_{1}$ as $c_{11}$, so an open set $(c_{00}, c_{01}) \cup (c_{10}, c_{11})$ is still not covered, and proceed further in this fashion.
4. In this proof only, we will denote $\overline{d_{\overline{m1}}...d_{\overline{mn}}}$ as $d_{m1}...d_{mn}$, a sequence of 0s and 1s, where m (first sub-subscript) denotes ordinal number of open interval in a union $O^\prime_{n}$ and second sub-subscript ranges from 1 to n where n is a step at which we are trying to cover all open sets. Let $ O_{d_{m1}...d_{m(n-1)}} = (c_{d_{m1}...d_{m(n-1)}0}, c_{d_{m1}...d_{m(n-1)}1}) $ and $O^\prime_{n-1} = \bigcup\limits_{m=1; \{d_{m1}...d_{m(n-1)}:1 \leq k \leq n, d_{mk} \in \{0,1\}\}}^{2^{n-1}} O_{d_{m1}...d_{m(n-1)}}$. Since at $n$-th step we have to cover $2^{n-1}$ open intervals, we can take a union of disjoint closed sets $C^\prime_{n-1} = \bigcup\limits_{m=1; \{d_{m1}...d_{m(n-1)}:1 \leq k \leq n-1, d_{mk} \in \{0,1\}\}}^{2^{n-1}} C_{d_{m1}...d_{m(n-1)}}$ and take its intersection with open set $O^\prime_{n-1}$ so that $C_{d_{m1}...d_{m(n-1)}} \cap O_{d_{m1}...d_{m(n-1)}} \neq \emptyset$. This implies $\forall m, 1 \leq m \leq 2^{n-1}, \hspace{0.1cm} \exists c_{d_{m1}...d_{m(n-1)}01} = \inf(O_{d_{m1}...d_{m(n-1)}} \cap C^\prime_{n-1}) $ and $\exists c_{d_{m1}...d_{m(n-1)}10} = \sup(O_{d_{m1}...d_{m(n-1)}} \cap C^\prime_{n-1}) \mid (c_{d_{m1}...d_{m(n-1)}00},c_{d_{m1}...d_{m(n-1)}01}) \cup (c_{d_{m1}...d_{m(n-1)}10},c_{d_{m1}...d_{m(n-1)}11}) \subset (O_{d_{m1}...d_{m(n-1)}} \setminus C^\prime_{n-1}) $. $(c_{d_{m1}...d_{m(n-1)}00},c_{d_{m1}...d_{m(n-1)}01}) = (c_{d_{k1}...d_{k(n-1)}d_{kn}0},c_{d_{k1}...d_{k(n-1)}d_{kn}1}) = O_{d_{k1}...d_{k(n-1)}d_{kn}} \subset O^\prime_n$. Denote $O = \bigcup\limits_{m=0}^{\infty} O^\prime_{m}$, $C = \bigcup\limits_{n=0}^{\infty} C^\prime_{n}$, $E = \{c_{d_{m1}...d_{mn}d_{mn}d_{mn}...}: k \in \mathbb{N}, d_{mk} \in \{0,1\}\}$. Also please note that for any endpoint in $E$ with finite subscript, such as $c_{d_{m1}d_{m2}...d_{mn}}$, $c_{d_{m1}d_{m2}...d_{mn}} = c_{d_{m1}d_{m2}...d_{mn}d_{mn}d_{mn}...}$, i.e. its finite subscript is equivalent to its infinite subscript which has last digit of its finite subscript repeating infinitely.
5. Since closed sets must cover entire $[0,1], \lim_{n \to +\infty} \sum\limits_{m=1;\{d_{m1}...d_{m(n-1)}: d_{mk} \in \{0,1\}, 1 \leq k \leq n-1\}}^{2^{n-1}} |c_{d_{m1}...d_{m(n-1)}1}-c_{d_{m1}...d_{m(n-1)}0}| = 0$. The number of closed sets will be countable (can get an injection to $\mathbb{Q} \subset [0,1]$: $f(m,n)=(2m-1)/2^n$ for $m$-th closed set inserted at $n$-th step, $2m-1<2^n$).
6. This infinite union of disjoint closed sets will not cover the entire $[0,1]$. Take an intersection of nested open intervals $\bigcap\limits_{n=1; d_{q_{k}k} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{q_{1}1}...d_{q_{n}n}} \mid O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}d_{q_{n}n}} \subset O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}}$ and $\forall n_{1} \in \mathbb{N} \hspace{0.1cm} \exists n_{2} \in \mathbb{N}, n_{2} > n_{1} \mid d_{q_{n_{2}}n_{2}}=1-d_{q_{n_{1}}n_{1}}$ and take a sequence of points $(o_{d_{q_{1}1}...d_{q_{n}n}}), o_{d_{q_{1}1}...d_{q_{n}n}} \in O_{d_{q_{1}1}...d_{q_{n-1}n-1}}$. $(o_{d_{q_{1}1}...d_{q_{n}n}}) \rightarrow L \notin C$. Proof: assume $L \in (C \setminus E) \implies \exists e_{1}, e_{2} \in E \mid L \in (e_{1}, e_{2})$, $[e_{1}, e_{2}] \subset C \implies \exists \epsilon > 0 \mid B_{\epsilon}(L) \cap O = \emptyset \implies L \notin (C \setminus E) $. Assume $L=c_{d_{m_{1}1}...d_{m_{n}n}d_{m_{n+1}(n+1)}...} \in E \implies \exists N \in \mathbb{N} \mid \forall n,k \geq N, d_{m_{k}k}=d_{m_{n}n}$. Take a sequence $(o_{d_{m_{1}1}...d_{m_{n}n}})$, $o_{d_{m_{1}1}...d_{m_{n}n}} \in O_{d_{m_{1}1}...d_{m_{n-1}n-1}}$, $O_{m}=(O \supset O_{d_{m_{1}1}} \supset O_{d_{m_{1}1}d_{m_{2}2}} \supset ... \supset O_{d_{m_{1}1}...d_{m_{N-1}(N-1)}d_{m_{N}N}d_{m_{N+1}N}}...)$. $\exists N^\prime \geq N \mid d_{q_{N^\prime}N^\prime} = 1-d_{m_{N^\prime}N^\prime} \implies \exists \epsilon > 0 \mid \forall N^{\prime\prime} > N^\prime$, $|o_{d_{q_{1}1}...d_{q_{N^{\prime\prime}}N^{\prime\prime}}}-o_{d_{m_{1}1}...d_{m_{N^{\prime\prime}}N^{\prime\prime}}}| > \epsilon$ since $O_{d_{q_{1}1}...d_{q_{N^{\prime\prime}-1}N^{\prime\prime}-1}} \cap O_{d_{m_{1}1}...d_{m_{N^{\prime\prime}-1}N^{\prime\prime}-1}} = \emptyset$. For the above $\epsilon/2$ there exist $N_{q} \in \mathbb{N} \mid \forall n \geq N_{q}$, $ |o_{q_{1}1q_{2}2...q_{n}n}-L| < \epsilon/2$, $N_{m} \in \mathbb{N} \mid \forall n \geq N_{m}$, $ |o_{m_{1}1m_{2}2...m_{n}n}-L| < \epsilon/2$. This implies $\exists M = max(N^{\prime\prime}, N_{q}, N_{m}) \mid \forall n \geq M$, $\epsilon < |o_{m_{1}1...m_{M}M}-o_{q_{1}1...q_{M}M}| = |o_{m_{1}1...m_{M}M}-L+L-o_{q_{1}1...q_{M}M}| \leq |o_{m_{1}1...m_{M}M}-L|+|L-o_{q_{1}1...q_{M}M}| < \epsilon/2+\epsilon/2 = \epsilon$ which is a contradiction $\implies L \notin (E \cup (C \setminus E)) \implies L \notin C$. Also, we can similarly prove that $L = \bigcap\limits_{n=1; d_{q_{k}k} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{q_{1}1}...d_{q_{n}n}} \mid O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}d_{q_{n}n}} \subset O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}}$ by assuming it doesn't belong to this intersection which will imply that there exists some $\epsilon > 0$ and some $N \in \mathbb{N} \mid \forall n \geq N, |o_{q_{1}1...q_{n}n}-L| > \epsilon$.
7. Denote $ N = \bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{m1}...d_{mn}} \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt}) = \{o_{d_{m1}d_{m2}...}: \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt}\}$ (by the above proof of intersection of nested open intervals being non-empty and converging to a point L in the intersection). This union amounts to a union of uncountably infinite number of points with each point having as subscript an infinite sequence consisting of 2 subsequences of 0's and 1's. Since $\forall L \in N, L \notin E \implies N \cap E = \emptyset$. We have to prove $N$ is not a countable union of disjoint closed sets. Proof:
   • Define a metric space $S = (N \cup E, d)$, $d$ is the Euclidian metric, $N \cap E = \emptyset$. It is complete which we prove next. $\forall n \in \mathbb{N}, O_{d_{m1}d_{m2}...d_{mn}} = (c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}) \implies N \cup E = (\bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} (c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}) \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt})) \cup \{c_{d_{m1}d_{m2}d_{m3}...d_{mn}d_{mn}d_{mn}...}: k \in \mathbb{N}, d_{mk} \in \{0,1\}\} = (\bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} [c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}] \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt})) = (\bigcap\limits_{n=1}^{\infty} (\bigcup\limits_{d_{mk} \in \{0,1\}, 1 \leq k \leq n} [c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}])$. It is closed since finite union of closed sets is closed, and intersection of an arbitrary number of closed sets is closed. Since $S$ is closed it is complete under the same Euclidian metric.
   • $N \in \mathbb{R}$ is defined as $N = \bigcap\limits_{n=1;\{d_{mk}: 1 \leq k \leq n, d_{mk} \in \{0,1\}\}}^{\infty} O_{d_{m1}d_{m2}...d_{mn}} = \bigcap\limits_{m=1}^{\infty} O_{m}^{\prime}$. Take some $m \in \mathbb{N}$, $O_{m}^{\prime}$ open in $\mathbb{R} \implies \forall x \in ((N \cup E) \cap O_{m}^{\prime}) \hspace{0.05cm} \exists \epsilon > 0 \mid B_{\epsilon}(x) \subset O_{m}^{\prime} \implies \forall y \in B_{\epsilon}(x), y \in (N \cup E)$ or $y \in (O_{m}^{\prime} \setminus (N \cup E)) \implies$ because $S$ consists of points in $N \cup E$, $N_{m}^{\prime} = (O_{m}^{\prime} \cap (N \cup E))$ is open in $S$. If $N \in \mathbb{R}, N=\bigcap\limits_{m=1}^{\infty} O_{m}^{\prime}=\bigcap\limits_{m=1}^{\infty} (O_{m}^{\prime} \cap (N \cup E))=\bigcap\limits_{m=1}^{\infty} N_{m}^{\prime}= N^{\prime}$, so $N^\prime$ (denoted as $N^{\prime}$ to indicate it refers to $S$) can be expressed as intersection of a countable number of open sets in $S$. Denote $E^{\prime}$ to indicate it refers to $S$. Assume $N^{\prime} = \bigcup\limits_{m=1}^{\infty} T_{m}, T_{m}$ closed $\implies (N^{\prime})^\complement = E^{\prime} = (\bigcup\limits_{m=1}^{\infty} T_{m})^\complement = \bigcap\limits_{m=1}^{\infty} E_{m}^{\prime}, \forall m \in \mathbb{N}, E_{m}^{\prime}$ open. Both $N$ and $E$ are dense in $S$ since $\forall n \in N$ and $\forall \epsilon > 0, \exists e \in E \mid e \in B_{\epsilon}(n) \implies$ since space $S$ is complete, $\overline{E} = (N \cup E) \implies \forall m \in \mathbb{N}, E_{m}^{\prime}$ is dense in $S$. Similarly, $\forall e \in E$ and $\forall \epsilon > 0, \exists n \in N \mid n \in B_{\epsilon}(e) \implies$ since space $S$ is complete, $\overline{N} = (E \cup N) \implies \forall m \in \mathbb{N}, N_{m}^{\prime}$ is dense in $S$. Therefore we have $E^{\prime} \cap N^{\prime} = (\bigcap\limits_{m=1}^{\infty} E_{m}^{\prime}) \cap (\bigcap\limits_{m=1}^{\infty} N_{m}^{\prime}) = \emptyset$, but by Baire Category Theorem, countable intersection of dense open sets is a proper subset of a complete metric space. So $N^{\prime}$ is not a union of a countable number of closed sets in $S$. Therefore $N$ is not a union of a countable number of closed sets in $\mathbb{R}$. Proof: assume $N = \bigcup\limits_{m=1}^{\infty} N_{m}, \forall m \in \mathbb{N} N_{m}$ closed. Then every $N_{m} \subset N \implies N_{m} \subset N^{\prime}$ in $S$. Assume $N_{m}^{\prime}$ is not closed in $S$, then $\exists (n_{k}) \rightarrow L \notin N_{m}^{\prime} \implies$ since $N_{m}$ and $N_{m}^{\prime}$ consist of the same elements, $(n_{k}) \rightarrow L \notin N_{m} \implies N_{m}$ is not closed which is a contradiction. Hence $N_{m}^{\prime}$ is closed in $S$ and $N^{\prime} = \bigcup\limits_{m=1}^{\infty} N_{m}^{\prime}$ is a union of a countable number of closed sets in $S$ which contradicts our proof above. Therefore the assumption is wrong and $N$ cannot be expressed as a union of a countable number of closed sets in $\mathbb{R}$. Therefore $N \cup C \cup B \cup A \neq [0,1]$ and $[0,1]$ is not a countable union of disjoint closed sets.