The idea of solving problems with absolute values is to break them into several cases where you can get rid of absolute values. Here there are four cases: (1) x >=0, y >=0; (2) x < 0, y >= 0; (3) x < 0, y < 0; (4) x >= 0, y < 0. By considering each of those cases, you rewrite |x| as either x or -x, and similarly for y.
I suggest you start with case (1), which has four subcases: (a) x - 2 >= 0, y - 2 >= 0; (b) x - 2 < 0, y - 2 >= 0; (c) x - 2 < 0, y - 2 < 0 and (d) x - 2 >= 0, y - 2 < 0. Find the set of points satisfying each of (a)—(d). In each of those subcases you can expand |x - 2| as either x - 2 or -(x - 2), and similarly for |y - 2|. After that, you don't have absolute values anymore, just equations of lines. Draw those lines, but only inside the area satisfying the corresponding subcase.
After you finish with case (1), note that if (x, y) satisfies the original equation, then so do (-x, y), (-x, -y) and (x, -y). Therefore, the polygon is symmetric with respect to the x- and y-axes. Thus, you can draw the parts corresponding to (2)—(4) just by reflecting the figure for (1) with respect to the axes.