A consequence of Anderson's inequality is that if $X$ and $Y$ are independent variables and $X$ is centered then $$P(\vert X+Y\vert\leq x)\leq P(\vert X\vert\leq x), x\geq 0. $$ Let $(X_1,X_2)$ be a random vector such that the r.v. are centered and normally distributed. It follows that $$P(\vert X_1\vert\leq x,\vert X_2\vert\leq x)=E\left(1_{\vert X_1\vert\leq x} P(\vert X_1-E(X_1\vert X_2)+E(X_1\vert X_2)\vert\leq x\vert X_2)\right).$$
Additionally $E(X_2(X_1 -E(X_1\vert X_2)))=0$ and thus $X_1 -E(X_1\vert X_2)$ and $X_2$ are independent.
According to a paper that I'm reading by putting the last two facts together with the inequality above one can proof that
$$P(\vert X_1\vert\leq x,\vert X_2\vert\leq x)\leq E\left(1_{\vert X_1\vert\leq x} P(\vert X_1-E(X_1\vert X_2)\vert\leq x)\right).$$
I would like to know how is it possible to deduce that.