Consider the set $T = \left\{ {x \in \left[{a \,.\,.\, b}\right]: \left[{a \,.\,.\, x}\right] \text { is covered by a finite subset of } S}\right\}$. === Step 1: $a \in T$ === It is demonstrated that the number $a$ is an element of $T$. Consider the interval $\left[ {a \,.\,.\, a} \right]$. $S$ covers $\left[{a \,.\,.\, a}\right]$ as it covers $\left[{a \,.\,.\, b}\right]$, a superset of $\left[{a \,.\,.\, a}\right]$. Since $S$ covers $\left[{a \,.\,.\, a}\right]$, there is a set $O$ in $S$ that contains $a$. We observe: :$\left\{ {O}\right\}$ is a subset of $S$ :$\left\{ {O}\right\}$ is finite as it contains one element :$\left\{ {O}\right\}$ covers $\left[{a \,.\,.\, a}\right]$ By the definition of $T$, $a$ is an element of $T$. === Step 2: $T$ has Supremum === It is demonstrated that $T$ has a supremum. We know that $a \in T$. Therefore, $T$ is non-empty. Also, $T$ is bounded as $T$ is a subset of $\left[{a \,.\,.\, b}\right]$ which is bounded. From these two properties of $T$, it follows by the continuum property of the real numbers that $T$ has a supremum, which will be denoted $c$. === Step 3: $c \in \left[{a \,.\,.\, b}\right]$ === It is demonstrated that the number $c$ is an element of $\left[{a \,.\,.\, b}\right]$. Every element of $T$ is less than or equal to $b$. So, $b$ is an upper bound for $T$. Therefore, $c \le b$ as $c$ is the least upper bound of T. Also, $c \ge a$ because $a \in T$ and $c$ is an upper bound for $T$. Accordingly, $c \in \left[{a \,.\,.\, b}\right]$. === Step 4: Finite Cover of $\left[{a \,.\,.\, y}\right]$ === It is demonstrated that there exists a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$ where $y > c$. We know that $S$ covers $\left[ {a \,.\,.\, b} \right]$. Also, $c \in \left[{a \,.\,.\, b}\right]$. From these two facts follows that there is a set $O_c$ in $S$ that contains $c$. An $\epsilon \in \R_{>0}$ exists such that $\left({c - \epsilon \,.\,.\, c + \epsilon}\right) \subset O_c$ as $O_c$ is open. A number $x \in T$ exists in $\left({c - \epsilon \,.\,.\, c} \right)$ as $c$ is the least upper bound of $T$. By the definition of $T$, a finite subset $S_x$ of $S$ exists such that $S_x$ covers $\left[{a \,.\,.\, x}\right]$ as $x \in T$. Let $y$ be a number in $\left({c \,.\,.\, c + \epsilon}\right)$. We observe that $y > c$. We have that $\left\{{O_c}\right\}$ covers $\left({c - \epsilon \,.\,.\, c + \epsilon}\right)$ as $\left({c - \epsilon \,.\,.\, c + \epsilon}\right) \subset O_c$. Therefore, $\left\{{O_c}\right\}$ covers $\left[{x \,.\,.\, y}\right]$ as $\left[{x \,.\,.\, y}\right] \subset \left({c - \epsilon \,.\,.\, c + \epsilon}\right)$. We observe the collection $S_x \cup \left\{{O_c}\right\}$: : $S_x \cup \left\{{O_c}\right\}$ is a subset of $S$ : $S_x \cup \left\{{O_c}\right\}$ is finite : $S_x \cup \left\{{O_c}\right\}$ covers $\left[{a \,.\,.\, y}\right]$, the union of $\left[{a \,.\,.\, x}\right]$ and $\left[{x \,.\,.\, y}\right]$ In other words, $S_x \cup \left\{ {O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$. === Step 5: $y > b$ === It is demonstrated that $y > b$. We know that $y > c$. Therefore, $y \notin T$ as $c$ is an upper bound for $T$. Let us find out exactly why $y \notin T$. There are two requirements for $y$ to be in $T$: : $\left[{a \,.\,.\, y}\right]$ is covered by a finite subset of $S$ : $y \in \left[{a \,.\,.\, b}\right]$ The first requirement is satisfied as $S_x \cup \left\{{O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$. Therefore, the second requirement is not satisfied as this is the last requirement and the only possibility to satisfy $y \notin T$. Accordingly, $y \notin \left[{a \,.\,.\, b}\right]$. Furthermore: {{begin-eqn}} {{eqn | l = y       | o = >       | r = c }} {{eqn | ll = \iff       | l = y       | o = >       | r = c \land c \ge a       | c = as $c \ge a$ since $c \in \left[{a \,.\,.\, b}\right]$ }} {{eqn | ll = \implies       | l = y       | o = >       | r = a }} {{eqn | ll = \iff       | l = y       | o = >       | r = a \land y∉[a..b]       | c = as $y \notin \left[{a \,.\,.\, b}\right]$ }} {{eqn | ll = \iff       | l = y       | o = >       | r = b }} {{end-eqn}} === Step 6: Finite Cover of $\left[{a \,.\,.\, b}\right]$ === It is demonstrated that there exists a finite subset of $S$ covering $\left[{a \,.\,.\, b}\right]$. We know that $S_x \cup \left\{{O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$. Also, $\left[{a \,.\,.\, b}\right]$ is a subset of $\left[{a \,.\,.\, y}\right]$ as $y > b$. Therefore, $S_x \cup \left\{{O_c}\right\}$ covers $\left[{a \,.\,.\, b}\right]$. So, $\left[{a \,.\,.\, b}\right]$ is covered by a finite subset of $S$. {{qed}} {{namedfor|Heinrich Eduard Heine|name2 = Émile Borel}}