1. Write the distances from the given lines

For a point $P(x,y)$, distance from the line $ax+by+c=0$ is $d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}.$

So,

• from $x-y=0$, $d_1=\frac{|x-y|}{\sqrt{2}}$
• from $x+y=0$, $d_2=\frac{|x+y|}{\sqrt{2}}.$

Since $P$ lies in the first quadrant, we have $x\ge 0, y\ge 0$, hence $x+y\ge 0 \implies |x+y|=x+y.$ Thus, $d_1+d_2=\frac{|x-y|+x+y}{\sqrt{2}}.$

We need $2\le \frac{|x-y|+x+y}{\sqrt{2}}\le 4.$ Multiplying by $\sqrt{2}$, $2\sqrt{2}\le |x-y|+x+y\le 4\sqrt{2}.$

2. Simplify $|x-y|+x+y$ in the first quadrant

Use the identity $|x-y|+x+y=2\max(x,y).$ Indeed:

• if $x\ge y$, then $|x-y|=x-y$, so $|x-y|+x+y=(x-y)+x+y=2x;$
• if $y\ge x$, then $|x-y|=y-x$, so $|x-y|+x+y=(y-x)+x+y=2y.$

Hence the condition becomes $2\sqrt{2}\le 2\max(x,y)\le 4\sqrt{2},$ so $\sqrt{2}\le \max(x,y)\le 2\sqrt{2}.$

3. Interpret the region geometrically

In the first quadrant, the set $\max(x,y)\le a$ is the square $0\le x\le a,\quad 0\le y\le a.$

Therefore, $\sqrt{2}\le \max(x,y)\le 2\sqrt{2}$ represents the region inside the square of side $2\sqrt{2}$ and outside the square of side $\sqrt{2}$, both in the first quadrant.

So area is $A=(2\sqrt{2})^2-(\sqrt{2})^2=8-2=6.$

4. Final answer

$\boxed{6}$

This matches the stored correct answer.