Prove that $ |x-a|I am not sure if I am doing this correctly,
my forward proof is:
Since $|x-a|$ is an absolute value function, it can be defined as $|x-a|:= \max\{(x-a),-(x-a)\}$ .
Let $S={(x-a) ∈ ℝ}$ and $b ∈ S, b>0,$
then $ x-a>b 
     x>b+a $

 2.-(x-a)>b

   -x+a>b

   -x>b-a

    x<a-b

which prove that x ∈ (a − b, a + b)
Am I doing this correctly, and how can I prove the backward?
Edit:
What I am thinking about the backward equation:
a-b < x , then a-b-x<0,  next -(a-b-x)>0,  so -(a-b-x)>(a-b-x) -->max = a-b-x
a+b>x , then a+b-x<0,  next -(a+b-x)>0,  so -(a+b-x)>(a-b-x) --> max= a+b-x
then I am stuck.

HINT
Assuming that $b > 0$, we have:
\begin{align*}
|x-a| < b \Longleftrightarrow |x-a|^{2} < b^{2} \Longleftrightarrow (x-a)^{2} - b^{2} < 0 \Longleftrightarrow (x - a - b)(x - a + b) < 0
\end{align*}
