Theorem: Lindelöf's Lemma/Lemma 1

Let $C$ be a set of open real sets.
Then there is a countable subset $D$ of $C$ such that:
:$\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$

Proof:
=== Lemma $2$ ===
{{:Lindelöf's Lemma/Lemma 2}}{{qed|lemma}}
Let $\ds U = \bigcup_{O \mathop \in C} O$.
Let $x$ be an arbitrary point in $U$.
Since $U$ is the union of the sets in $C$, the point $x$ belongs to a set in $C$.
Name such a set $O_x$.
Since $O_x$ is open, $O_x$ contains an open interval $I_x$ that contains $x$.
By Between two Real Numbers exists Rational Number, a rational number exists between the left hand endpoint of $I_x$ and $x$.
Also, a rational number exists between $x$ and the right hand endpoint of $I_x$.
Form an open interval $R_x$ that has two such rational numbers as endpoints.
All in all:
{{begin-eqn}}
{{eqn | l = x \in R_x
      | o = \subset
      | r = I_x \subset O_x \in C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \in R_x
      | o = \subset
      | r =  O_x \in C
}}
{{end-eqn}}
By {{Lemma|Lindelöf's Lemma|2}}, $\set {R_x: x \in U}$ is countable as $\set {R_x: x \in U}$ is a set of open intervals with rational numbers as endpoints.
By Countable Set equals Range of Sequence, the countability of $\set {R_x: x \in U}$ means that there exists a sequence $\sequence {R^i}_{i \mathop \in N}$ where:
:$N$ is a subset of $\N$
:$\set {R_x: x \in U}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.
From this follows:
:$\set {R_x: x \in U} = \set {R^i: i \in N}$ as $\set {R^i: i \in N}$ equals the range of $\sequence {R^i}_{i \mathop \in N}$.
Two sequences that differ only by one of them having duplicates, have the same range.
Therefore, it is possible to require that $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.
Now, let $i$ be an arbitrary natural number in $N$.
Let $R^i$ be an element in $\set {R^i: i \in N}$.
There is an $x$ in $U$ such that:
:$R_x = R^i$ as $\set {R_x: x \in U} = \set {R^i: i \in N}$
Also, we know that a set $O_x$ in $C$ exists such that:
:$R_x \subset O_x$
The uniqueness of the elements of N makes it possible to define a mapping $\chi$ that sends $i$ to $x$.
This allows us to define, for every $i$ in $N$:
:$O^i = O_x$ where $x = \map \chi i$
We find, for every $i$ in $N$ and $x = \map \chi i$:
:$O^i \in C$ as $O_x \in C$
:$R^i \subset O^i$ as $R_x \subset O_x$ and $R_x = R^i$ and $O_x = O^i$
Every $R^i$ where $i \in N$ is uniquely determined by $i$ as $\sequence {R^i}_{i \mathop \in N}$ lacks duplicates.
Therefore, it is possible to define a mapping from $\set {R^i: i \in N}$ to $\set {O^i: i \in N}$ that sends $R^i$ to $O^i$ for every $i$ in $N$.
Image of countable set under mapping is countable.
Therefore, $\set {O^i: i \in N}$ is countable as $\set {R^i: i \in N}$ is countable.
$\set {O^i: i \in N}$ is a subset of $C$ as $O^i \in C$ for every $i \in N$.
Therefore, $\set {O^i: i \in N}$ is a countable subset of $C$.
We find by focusing on $R_x$ for an $x$ in $U$:
{{begin-eqn}}
{{eqn | l = x  
      | o = \in
      | r = R_x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \set x
      | o = \subset
      | r = R_x
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{x \mathop \in U} \set x
      | o = \subseteq
      | r = \bigcup_{x \mathop \in U} R_x
}}
{{eqn | ll= \leadsto
      | l = U
      | o = \subseteq
      | r = \bigcup_{x \mathop \in U} R_x
}}
{{end-eqn}}
So, $\set {R_x: x \in U}$ covers $U$.
We find by focusing on $O^i$ for an $i$ in $N$:
{{begin-eqn}}
{{eqn | l = R^i
      | o = \subset
      | r = O^i
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{i \mathop \in N} R^i
      | o = \subseteq
      | r = \bigcup_{i \mathop \in N} O^i
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{x \mathop \in U} R_x
      | o = \subseteq
      | r = \bigcup_{i \mathop \in N} O^i
      | c = as $\set {R_x: x \in U} = \set {R^i: i \in N}$
}}
{{eqn | ll= \leadsto
      | l = U
      | o = \subseteq
      | r = \bigcup_{x \mathop \in U} R_x \subseteq \bigcup_{i \mathop \in N} O^i
      | c = as $\ds U \subseteq \bigcup_{x \mathop \in U} R_x$
}}
{{eqn | ll= \leadsto
      | l = U
      | o = \subseteq
      | r = \bigcup_{i \mathop \in N} O^i
}}
{{eqn | ll= \leadsto
      | l = U
      | o = \subseteq
      | r = \bigcup_{i \mathop \in N} O^i \subseteq U
      | c = as $\ds \bigcup_{i \mathop \in N} O^i \subseteq U$ is true since every $O_i \in C$
}}
{{eqn | ll= \leadsto
      | l = U
      | r = \bigcup_{i \mathop \in N} O^i
}}
{{end-eqn}}
So, $D = \set {O^i: i \in N}$ satisfies the proposition of the theorem.
{{qed}}