=== Lemma === {{:Lindelöf's Lemma/Lemma/Lemma|Lemma}} {{qed|lemma}} Let $U = \displaystyle \bigcup_{O \mathop \in C} O$. 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 = \implies       | l = x \in R_x       | o = \subset       | r =  O_x \in C }} {{end-eqn}} By Lemma, $\left\{{R_x: x \in U}\right\}$ is countable as $\left\{{R_x: x \in U}\right\}$ is a set of open intervals with rational numbers as endpoints. By Countable Set equals Range of Sequence, the countability of $\left\{{R_x: x \in U}\right\}$ means that there exists a sequence $\left\langle{R^i}\right\rangle_{i \mathop \in N}$ where: : $N$ is a subset of $\N$ : $\left\{{R_x: x \in U}\right\}$ equals the range of $\left\langle{R^i}\right\rangle_{i \mathop \in N}$. From this follows: : $\left\{{R_x: x \in U}\right\} = \left\{{R^i: i \in N}\right\}$ as $\left\{{R^i: i \in N}\right\}$ equals the range of $\left\langle{R^i}\right\rangle_{i \mathop \in N}$. Two sequences that differ only by one of them having duplicates, have the same range. Therefore, it is possible to require that $\left\langle{R^i}\right\rangle_{i \mathop \in N}$ lacks duplicates. Now, let $i$ be an arbitrary natural number in $N$. Let $R^i$ be an element in $\left\{{R^i: i \in N}\right\}$. There is an $x$ in $U$ such that: : $R_x = R^i$ as $\left\{{R_x: x \in U}\right\} = \left\{{R^i: i \in N}\right\}$ 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 $x = \chi \left({i}\right)$ We find, for every $i$ in $N$ and $x = \chi \left({i}\right)$: : $O^i \in C$ 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 $\left\langle{R^i}\right\rangle_{i \mathop \in N}$ lacks duplicates. Therefore, it is possible to define a mapping from $\left\{{R^i: i \in N}\right\}$ to $\left\{{O^i: i \in N}\right\}$ that sends $R^i$ to $O^i$ for every $i$ in $N$. Image of countable set under mapping is countable. Therefore, $\left\{ {O^i: i \in N} \right\}$ is countable as $\left\{ {R^i: i \in N} \right\}$ is countable. $\left\{{O^i: i \in N}\right\}$ is a subset of $C$ as $O^i \in C$ for every $i \in N$. Therefore, $\left\{{O^i: i \in N}\right\}$ 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 = \iff       | l = \left\{ {x}\right\}       | o = \subset       | r = R_x }} {{eqn | ll = \implies       | l = \bigcup_{x \mathop \in U} \left\{ {x} \right\}       | o = \subseteq       | r = \bigcup_{x \mathop \in U} R_x }} {{eqn | ll = \iff       | l = U       | o = \subseteq       | r = \bigcup_{x \mathop \in U} R_x }} {{end-eqn}} So, $\left\{{R_x: x \in U}\right\}$ 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 = \implies       | l = \bigcup_{i \mathop \in N} R^i       | o = \subseteq       | r = \bigcup_{i \mathop \in N} O^i }} {{eqn | ll = \iff       | l = \bigcup_{x \mathop \in U} R_x       | o = \subseteq       | r = \bigcup_{i \mathop \in N} O^i       | c = as $\left\{ {R_x: x \in U}\right\}$ = $\left\{ {R^i: i \in N}\right\}$ }} {{eqn | ll = \iff       | l = U       | o = \subseteq       | r = \bigcup_{x \mathop \in U} R_x \subseteq \bigcup_{i \mathop \in N} O^i       | c = as $U \subseteq \displaystyle \bigcup_{x \mathop \in U} R_x$ }} {{eqn | ll = \implies       | l = U       | o = \subseteq       | r = \bigcup_{i \mathop \in N} O^i }} {{eqn | ll = \iff       | l = U       | o = \subseteq       | r = \bigcup_{i \mathop \in N} O^i \subseteq U       | c = as $\displaystyle \bigcup_{i \mathop \in N} O^i \subseteq U$ is true since every $O_i \in C$ }} {{eqn | ll = \iff       | l = U       | r = \bigcup_{i \mathop \in N} O^i }} {{end-eqn}} So, $D = \left\{{O^i: i \in N}\right\}$ satisfies the proposition of the theorem. {{qed}}