How to solve $|x-1|+|x+1|<2$-- my method does not work for this one I've just successfully solved the following exercise:
Find $x$ such that $|x-1| + |x-2| >1$.
I did it by making four cases depending on whether the two absolute value expressions are positive or negative.
I moved on the next exercise expecting to solve it using the same method but what I got is completely wrong (not even close!?).

Please could someone explain to me why my method does not work here?

Here is the exercise and what I did:
Find $x$ with $|x-1| + |x+1| < 2$.
We have $|x-1| + |x+1| < 2$ if and only if one of the four cases is satisfied:
Case 1: $x - 1 + x + 1 < 2$ if and only if $2x < 2$ if and only if $x < 1$.
Case 2: $x-1 - x - 1 < 2$ if and only if $0 < 4$ which is always true. This case would imply that the inequality is true for all $x \in \mathbb R$.
Case 3: $-x + 1 + x + 1 < 2$ if and only if $2<2$ which is always false.
Case 4: $-x + 1 - x - 1 < 2$ if and only if $x > -1$.
 A: The flaw in your reasoning has to do with the fact that when you consider each case, you don't consider the conditions on $x$ for which that case applies.  For example, if you write in Case 2 that $x-1-x-1 < 2$, under what conditions is it true that $|x-1| + |x+1| = (x-1) - (x+1)$?  You conclude that $0 < 4$, so that this case is always true.  But it is not:  the case is valid only when $|x-1| = x-1$, and $|x+1| = -(x+1)$.  When does this happen?  The first condition happens when $x-1 \ge 0$, and the second condition happens when $x+1 < 0$.  Thus, Case 2 is valid if and only if $x \ge 1$ and $x < -1$.  But this is a contradiction; no such $x$ is simultaneously at least $1$ and less than $-1$.  So the conclusion you've drawn from Case 2 is impossible.
This suggests that in order to solve the inequality properly, you need only consider those intervals for which $x-1$ and/or $x+1$ changes sign, or equivalently, the partitions of the real line at $x = -1$ and $x = 1$:  $$x \in (-\infty, -1), \quad x \in [-1, 1), \quad x \in [1, \infty).$$
On the first of these intervals, $x-1 < 0$, hence $|x-1| = -(x-1)$; and $x+1 < 0$, hence $|x+1| = -(x+1)$.
On the second of these intervals, $x-1 < 0$, but $x+1 \ge 0$.
And on the third of these intervals, both $x-1 \ge 0$ and $x+1 \ge 0$.
A: You need to consider 3 regions not 4 cases (considering 2 roots for each absolute value expression) $x<-1$ , $-1\leq x\leq 1$, $x> 1$.
