I came up with what I think is the solution to exercise 11. (v) on chapter 1 of the third edition of book Calculus by Michael Spivak.
Find all numbers $x$ for which $|x - 1| + |x + 1| < 2$.
Solution. Three cases needed consideration.
First, in case $x < -1$, $|x - 1| = -(x - 1)$ while $|x + 1| = - (x + 1)$, so $|x - 1| + |x + 1| = -(x - 1) - (x + 1) < 2$ giving $x > -1$
Secondly, if $-1 \le x < 1$ then $|x - 1| = -(x - 1)$ and $|x + 1| = x + 1$, so $|x - 1| + |x + 1| = -(x - 1) + (x + 1) < 2$. But $2 < 2$ is a contradiction, so $\varnothing$ is the set of solution for the second case.
Lastly, in case $x \ge 1$, $|x - 1| = x - 1$ and $|x + 1| = x + 1$, so $|x - 1| + |x + 1| = (x - 1) + (x + 1) < 2$ giving $x < 1$.
The set of solutions for the inequality is supposedly $(-1, 1)$. But the second case indicates that every element in the interval fails the elementhood test, so the set of solutions is $\varnothing$.
I have two questions regarding my solution. First, is my approach to the solution correct? Secondly, if it is correct then is there a more concise method showing that the set of solutions is $\varnothing$?