I'm going to switch your variables in the statement so that the discussion uses them in a more conventional sense of studying $Y$ conditional on the values of $X$. So, the objective is to show that
$$\Pr(Y\in J\mid X\in I) \ge \Pr(Y\in J).\tag{1}$$
Think of this in terms of regression:
We will exploit the fact that the distribution of $Y$ conditional on $X=x$ is a standard distribution $F$ (in this case Normal with mean zero and some given variance $\sigma^2$) whose location has been shifted to $f(x)$ for some function $f$ (the regression function). By letting $J-f(x)$ stand for all values $y-f(x)$ where $y\in J$,this means $$\Pr(Y\in J\mid X=x) = {\Pr}(Y-f(x)\in J-f(x)\mid X=x)=\int_{J-f(x)} \mathrm{d}F(y).\tag{2}$$
Let's suppose that intervals centered around $0$ are maximum-probability intervals for $F$: among all intervals of the same width, these have the largest probability. That's clearly the case for symmetric unimodal distributions like the Normal. This can be written $$\int_J \mathrm{d}F(y) \ge \int_{J-a} \mathrm{d}F(y)\tag{3}$$ for any number $a$.
Assume $\Pr(X\in I) \gt 0$. (The other case is trivial to prove). This allows us to write
$$\Pr(Y\in J\mid X\in I) = \frac{\Pr(Y\in J\text{ and } X\in I)}{\Pr(X\in I)}.\tag{4}$$
Now, letting $G$ stand for the distribution of $X$, we can compare the two sides of $(1)$ by means of $(2)$, applying $(3)$ with $a=f(x)$, and simplifying the resulting double integral:
$$\eqalign{
\Pr(Y\in J\text{ and } X\in I) &= \int_I \Pr(Y\in J\mid X=x) \mathrm{d}G(x) \\
&=\int_I\int_{J-f(x)}\mathrm{d}F(y)\ \mathrm{d}G(x) \\
&\le \int_I\int_J \mathrm{d}F(y)\mathrm{d}G(x) \\
&= \int_J \mathrm{d}F(y) \int_I\mathrm{d}G(x) \\
&=\Pr(Y\in J)\Pr(X\in I).
}$$
In light of $(4)$, dividing both sides by $\Pr(X\in I)$ produces $(1)$, QED.