jmc

First, assume that $x\ge 0$ and $y\ge 0.$ If $y\ge x,$ then $$|x+y|+|x-y|=x+y+y-x=2y\le 4,$$so $y\le 2.$ If $y<x,$ then $$|x+y|+|x-y|=x+y+x-y=2x\le 4,$$so $x\le 2.$

Thus, the portion of the graph in the first quadrant is as follows:



Now, suppose $(a,b)$ satisfies $|x+y|+|x-y|\le 4,$ so $$|a+b|+|a-b|\le 4.$$If we plug in $x=a$ and $y=-b,$ then $$|x+y|+|x-y|=|a-b|+|a+b|\le 4.$$This means if $(a,b)$ is a point in the region, so is $(a,-b).$ Therefore, the region is symmetric around the $x$-axis.

Similarly, if we plug in $x=-a$ and $y=b,$ then $$|x+y|+|x-y|=|-a+b|+|-a-b|=|a-b|+|a+b|\le 4.$$This means $(-a,b)$ is also a point in the region. Therefore, the region is symmetric around the $y$-axis.

We conclude that the whole region is a square with side length 4.



Hence, its area is $\boxed{16}.$