We are given that $C$ is a set of open real sets that covers $F$. In other words, $C$ is an open cover of $F$. We need to show that there is a finite subset of $C$ that covers $F$. In other words, we need to show that $C$ has a finite subcover. Let $F_o$ be the complement of $F$ in $\R$. By the definition of closed real set, $F_o$ is open as $F$ is closed. === Step 1: $C^*$ is an Open Cover of $\left[{a \,.\,.\, b}\right]$ === It is demonstrated that $C^*$ is an open cover of $\left[{a \,.\,.\, b}\right]$. Since $F$ is bounded, $F$ is contained in a closed and bounded interval $\left[{a \,.\,.\, b}\right]$ where $a, b \in \R$. Define $C^* = C \cup \left\{ {F_o}\right\}$. Like $C$, $C^*$ is a set of open real sets as $F_o$ is open. $C^*$ covers $F \cup F_o$ as $C$ covers $F$ and $\left\{ {F_o} \right\}$ covers $F_o$. $F_o \cup F$ equals $\R$ as $F_o$ is the complement of $F$ in $\R$. So $C^*$ covers $\R$. Furthermore, $C^*$ is an open cover of $\left[{a \,.\,.\, b}\right]$ as $\left[{a \,.\,.\, b}\right]$ is a subset of $\R$. === Step 2: $C^*$ has a Finite Subcover === It is demonstrated that $C^*$ has a finite subcover $C^*_f$. $C^*$ is an open cover of the closed and bounded interval $\left[{a \,.\,.\, b}\right]$. Therefore, by Open Cover of Closed and Bounded Real Interval has Finite Subcover, $C^*$ has a finite subcover $C^*_f$. === Step 3: $C$ has a Finite Subcover === It is demonstrated that $C$ has a finite subcover $C_f$. Note that $F_o$ is the only element in $C^*$ that is not an element of $C$. Therefore, $F_o$ is the only possible element in $C^*_f$ that is not an element of $C$ as $C^*_f$ is a subset of $C^*$. This means that $C^*_f$ \ $\left\{ {F_o} \right\}$ is a subset of $C$. Define $C_f$ = $C^*_f$ \ $\left\{ {F_o} \right\}$. According to the reasoning above, $C_f$ is a subset of $C$. Also, $C_f$ is finite as $C^*_f$ is finite. What remains is to show that $C_f$ covers $F$. We have: {{begin-eqn}} {{eqn | o = \implies       | l = F \subseteq \left[ {a \,.\,.\, b} \right]       | r = F \subseteq \left[ {a \,.\,.\, b} \right] \subseteq \bigcup_{O \mathop \in C^*_f} O       | c = as $C^*_f$ is a cover of $\left[ {a \,.\,.\, b} \right]$ }} {{eqn | o = \implies       | r = F \subseteq \bigcup_{O \mathop \in C^*_f} O }} {{eqn | o = \implies       | r = F \cap \complement F_o \subseteq (\bigcup_{O \mathop \in C^*_f} O) \cap \complement F_o       | c = Set Intersection Preserves Subsets }} {{eqn | r = F \cap \complement \complement F \subseteq (\bigcup_{O \mathop \in C^*_f} O) \cap \complement F_o       | c = as $F_o = \complement F$, the complement of $F$ in $\R$ }} {{eqn | r = F \cap F \subseteq (\bigcup_{O \mathop \in C^*_f} O) \cap \complement F_o       | c = Relative Complement of Relative Complement }} {{eqn | r = F \subseteq (\bigcup_{O \mathop \in C^*_f} O) \cap \complement F_o       | c = Intersection is Idempotent }} {{end-eqn}} Furthermore, as $F \subseteq \left[{a \,.\,.\, b}\right]$: {{begin-eqn}} {{eqn | o = \subseteq       | l = F       | r = \left({\bigcup_{O \mathop \in C^*_f} O}\right) \cap \complement F_o }} {{eqn | o = \subseteq       | r = \left({\bigcup_{O \mathop \in C^*_f \cup \left\{ {F_o}\right\} } O}\right) \cap \complement F_o       | c = by $\displaystyle \left({\bigcup_{O \mathop \in C^*_f} O}\right) \subseteq \left({\bigcup_{O \mathop \in C^*_f \cup \left\{ {F_o}\right\} } O}\right)$ and Set Intersection Preserves Subsets }} {{eqn | r = \left({\bigcup_{O \mathop \in C^*_f \cup \left\{ {F_o}\right\} } O}\right) \setminus F_o       | c = as set intersection with complement equals set difference }} {{eqn | r = \left({\bigcup_{O \mathop \in \left({C^*_f \setminus \left\{ {F_o}\right\} \cup \left\{ {F_o}\right\} }\right)} O}\right) \setminus F_o       | c = by Set Difference Union Second Set is Union }} {{eqn | r = \left({\left({\bigcup_{O \mathop \in \left({C^*_f \setminus \left\{ {F_o}\right\} }\right)} O}\right) \cup F_o}\right) \setminus F_o       | c =  }} {{eqn | r = \left({\bigcup_{O \mathop \in \left({C^*_f \setminus \left\{ {F_o}\right\} }\right)} O}\right) \setminus F_o       | c = Set Difference with Union is Set Difference }} {{eqn | o = \subseteq       | r = \bigcup_{O \mathop \in \left({C^*_f \setminus \left\{ {F_o}\right\} }\right)} O       | c = Set Difference is Subset }} {{eqn | r = \left({\bigcup_{O \mathop \in C_f} O}\right)       | c = Definition of $C_f$ }} {{end-eqn}} Thus, $C_f$ covers $F$. This finishes the proof. {{qed}} 	 {{Namedfor|Heinrich Eduard Heine|name2 = Émile Borel}}