We consider 4 cases: Case 1$x\geq0$, $y\geq0$. We have: S1$2x+2y\leq2$, so $x+y\leq1$. S2$0+0\leq$, so $0\leq2$. We then have the area bounded by $x+y\leq1$, and the $x$ and $y$ axis, so we get $\frac{1\cdot1}{2}=\frac{1}{2}$. Case 2$x<0$, $y\geq0$. We have: S1$2y\leq2$, so $y\leq1$ S2$-2x\leq2$, so $x\geq-1$ We then have a 1x1 square, with an area of $1\cdot1=1$. Case 3$x<0$, $y<0$. We have: S1$0+0\leq2$, so $0\leq2$ S2$-2x-2y\leq2$, so $x+y\geq-1$ We then have a triangle with area $\frac{1\cdot1}{2}=\frac{1}{2}$. Case 4$x\geq0$, $y<0$. S1We have $2x\leq2$, so $x\leq0$ S2We have $-2y\leq2$, so $y\geq-1$ We then have a 1x1 square, which has area $1\cdot1=1$. Our total, then, is $\frac{1}{2}+1+\frac{1}{2}+1=\boxed{3}$

[asy][asy] size(8cm);import graph; path p=(-3,1)--(0,1)--(1,0)--(1,-3); path q=(-1,3)--(-1,0)--(0,-1)--(3,-1); fill(buildcycle(p,q),gray(.9)); draw(p^^q,blue+dashed); xlimits(-3,3);ylimits(-3,3);yaxis("$y$",RightTicks());xaxis("$x$",Ticks()); draw((-1,0)--(1,0)^^(0,1)--(0,-1)); [/asy][/asy]