Given $3$ points $A,B,C$ such that $A-B-C$ then the line segment $\overline{AC}= \overline {AB} \cup \overline{BC}$
My attempt so far at the proof, I know it must use mutual containment but feel my attempt is sloppy.
Let $U$ be the universal set where all the points lie
Let $\overline{AB}=\left\{X \in U: X=A,\hspace{2mm} \text{or} \hspace{2mm} A-X-B \hspace{2mm}\text{or} \hspace{2mm} X=B \right\}$
Let $\overline{BC}=\left\{X \in U: X=B,\hspace{2mm} \text{or} \hspace{2mm} B-X-C \hspace{2mm}\text{or} \hspace{2mm} X=C \right\}$
Let $\overline{AC}=\left\{X \in U: X=A,\hspace{2mm} \text{or} \hspace{2mm} A-X-C \hspace{2mm}\text{or} \hspace{2mm} X=C \right\}$
By the ruler postulate consider the function $f:\overleftrightarrow{AB}=\overleftrightarrow{BC}=\overleftrightarrow{AC} \rightarrow \mathbb{R}$ such that $f(A)=a, f(B)=b, f(C)=c$. Then we may assume that $a<b<c$. Let $X \in \overline{AC}$ such that $f(X)=x$ then we have that $a\leq x \leq b$ or that $b \leq x \leq c$. Hence $X \in \overline {AB} \cup \overline{BC}$, therefore we have that $\overline{AC} \subseteq\overline {AB} \cup \overline{BC}$. Now let $X \in \overline {AB} \cup \overline{BC}$ such that $f(X)=x$ then we have that $a\leq x \leq b$ or that $b \leq x \leq c$ in either case $a \leq x \leq c$ since by assumption we have that $a<b<c$, thus $X \in \overline{AC}$. Hence we have that $\overline {AB} \cup \overline{BC}\subseteq \overline{AC}$. Therefore by mutual containment $\overline{AC}= \overline {AB} \cup \overline{BC}$
Any help to clean this proof up or make the argument more clear would be appreciated.