jmc

We'll use this helpful result a few times: for any real numbers $a$ and $b$, the graph of $$|x-a|+|y-b|=1$$is a "diamond": a square of side length $\sqrt{2}$ centered at $(a,b)$ whose sides form angles of $45^{\circ }$ with the axes. (To see this, first draw the graph of $|x|+|y|=1$. Then, the graph of $|x-a|+|y-b|=1$ is just the result of translating in the $x$-direction by $a$, and then in the $y$-direction by $b$.)

Since the given equation only involves $|x|$ and $|y|$, it is symmetric about the two axes. That is, it is sufficient to consider only the first quadrant, and then multiply our answer by $4$ to account for all four quadrants. So, assume $x,y\ge 0$.Then the equation becomes $$\left|\right|x-2|-1|+\left|\right|y-2|-1|=1.$$Seeing $|x-2|$ and $|y-2|$, we take cases on the values of $x$ and $y$ relative to $2$:

If $0\le x,y\le 2$, then the given equation becomes $$|(2-x)-1|+|(2-y)-1|=1⟹|1-x|+|1-y|=1.$$This is the equation of the standard diamond centered at $(1,1)$, which is completely contained in the region $0\le x,y\le 2$. If $0\le x\le 2\le y$, then the given equation becomes $$|(2-x)-1|+|(y-2)-1|=1⟹|1-x|+|y-3|=1.$$This is the equation of the standard diamond centered at $(1,3)$, which is again contained in the correct region. If $0\le y\le 2\le x$, then we get the standard diamond centered at $(3,1)$, as in the last case. If $2\le x,y$, then the given equation becomes $$|(x-2)-1|+|(y-2)-1|=1⟹|x-3|+|y-3|=1.$$This is the equation of the standard diamond centered at $(3,3)$, which is again contained in the region $2\le x,y$.

Thus, the graph of the given equation in the first quadrant consists of four standard diamonds, so the graph of the given equation in the whole plane consists of $4\cdot 4=16$ standard diamonds. These diamonds do not overlap, and each one has perimeter $4\sqrt{2}$. So, the overall length of the lines that make up the graph is $16\cdot 4\sqrt{2}=\boxed{64\sqrt{2}}$.

Below is the whole graph of the equation (tick marks are at $x,y=\pm 1,\pm 2,\dots$).