$\because \left|x-1\right|+\left|x+1\right|\left\{\begin{array}{cc}-2x,& x<-1\\ -2,& -1\le x\le 1\\ 2x,& x>1\end{array}\right.$

For, $x<-1$

$-2x<4\Rightarrow x>-2$

$\Rightarrow x\in \left(-2,-1\right) ... \left(\mathrm{i}\right)$

For, $x\in \left[-1,1\right]$

$-2<4$

$\Rightarrow x\in \left[-1,1\right] ... \left(\mathrm{ii}\right)$
and for $x>1$

$2x<4\Rightarrow x<2$

$\Rightarrow x\in \left(1,2\right) ... \left(\mathrm{iii}\right)$
From Equations. $\left(\mathrm{i}\right), \left(\mathrm{ii}\right)$ and $\left(\mathrm{iii}\right)$.

$x\in \left(-2,2\right)$

Alternate Method:

Given that,

$\left|x-1\right|+\left|x+1\right|<4$

By verifying the options:

From option $\left(1\right)$, 

Let $x=-1$

$\Rightarrow \left|-1-1\right|+\left|-1+1\right|=\left|-2\right|+\left|0\right|=2<4$

From option $\left(2\right)$, 

Let $x=-3$

$\Rightarrow \left|-3-1\right|+\left|-3+1\right|=\left|-4\right|+\left|-2\right|$

$=4+2$

$=6\nless 4$

From option $\left(3\right)$, 

Let $x=2$

$\Rightarrow \left|2-1\right|+\left|2+1\right|=1+3$

$=4\nless 4$

From option $\left(4\right)$, 

Let $x=1.5$

$\Rightarrow \left|1.5-1\right|+\left|1.5+1\right|=\left|-1.5\right|+\left|2.5\right|$

$=1.5+2.5$

$=4\nless 4$

$\therefore x=(-2,2)$.