We view the problem geometrically, or equivalently in terms of motion.
Suppose that we are travelling on the $x$-axis, and are now at the point $(x,0)$. Then $\sqrt{(x-1)^2}$ is our distance from the point $(1,0)$. Similarly, $\sqrt{(x+1)^2}=\sqrt{(x-(-1))^2}$ is our distance from the point $(-1,0)$.
It follows that $\sqrt{(x+1)^2}+\sqrt{(x-1)^2}$ is the sum of our distances from $(-1,0)$ and $(1,0)$.
Suppose that $x$ is anywhere between $-1$ and $1$. Then it is clear that the sum of our distances from $(-1,0)$ and $(1,0)$ is $2$. Any small motion to the right decreases our distance from $1$, but increases our distance from $-1$ by the same amount.
Suppose that $x>1$. Then our distance from $(1,0)$ is $x-1$. Our distance from $(-1,0)$ is $2$ more than that, so it is $x+1$. The sum is $2x$. The sum of the distances is increasing at twice the rate that the distance from $(1,0)$ is increasing.
Suppose finally that $x<-1$. By symmetry, the sum of the distances is the same as the sum for $|x|$. By the preceding paragraph, this sum is $2|x|$, which could also be written as $-2x$.
Generalization: We could make a "word problem" which is answered by the above calculation. Adam lives at $(-1,0)$ and Beth lives at $(1,0)$. They want to meet at the point $(x,0)$. What is the sum of the distances they must travel?
The problem can be generalized. Suppose that we have $n$ people, who live respectively at $(a_1,0)$, $(a_2,0)$, $\dots$, $(a_n,0)$, where $a_1\le a_2\le\cdots\le a_n$. What is the sum of their distances from the point $(x,0)$? The analysis is not much more complicated than for $2$ people.