Isn't $\mathbb{E}[X^2|X\in[-b,b]]\leq \mathbb{E}[X^2]$ true? Let $X$ be a centered random variable such that $\mathrm{support}(X)\subseteq[-B,B]$, $B>0$, with discrete or continuous density $f$.
Now, consider an event $\xi=\{X\in [-b,b]\}$, $b\in[0,B)$, with $\mathbb{P}(\xi)\in(0,1)$. Then we have
$$\mathbb{E}[X^2|\xi]=\frac{1}{\mathbb{P}(\xi)}\int x^2\mathbb{1}_{[-b,b]}(x)f(x)\ \mathrm{d} x\leq \frac{1}{\mathbb{P}(\xi)}\int x^2f(x)\ \mathrm{d} x= \frac{1}{\mathbb{P}(\xi)}\mathbb{E}[X^2].$$
(Where the integral is meant in the Lebesgue- or Dirac- sense.)
However, intuitively, in this setting I would expect $\mathbb{E}[X^2|\xi]\leq \mathbb{E}[X^2]$.
So my questions are: If the latter result is true, how can it be proved? If it is wrong, can you provide a counter-example?  
 A: Yes, this is true, and it follows from a variant of the FKG inequality. However, it seems simplest to give a self-contained proof.
Proof of inequality.
Let $\mu$ be the law of $X$. The inequality you wish to prove is equivalent to
\begin{equation}
\int_{\mathbb R}x^2\ \mathbf{1}[-b,b]\ d\mu\leq \int_{\mathbb R}\mathbf{1}[-b,b]\ d\mu \int_{\mathbb R}x^2\ d\mu.\qquad (\star)
\end{equation}
Consider the integral
$$
I=\int_{\mathbb R} \int_{\mathbb R} \bigl(x^2-y^2\bigr)\bigl(\mathbf{1}[-b,b](x)-\mathbf{1}[-b,b](y)\bigr)\ d\mu(x)\ d\mu(y).
$$
We claim that $I\leq 0$. Indeed, the integrand is always less than or equal to $0$. This is because if $|x|> |y|$, the first factor is positive 
 and the second factor is either $0$ or $-1$, and if $|x|<|y|$, the first factor is negative while the second is either $0$ or $1$. Thus the integrand is at most $0$, so $I\leq 0$.
On the other hand, we claim that
$$I=2\cdot\text{LHS of }(\star)-2\cdot\text{RHS of }(\star).$$
Indeed, this follows by expanding out $I$ and using the fact that $\mu$ is a probability measure.
\begin{align*}
I&=\int_{\mathbb R}\int_{\mathbb R} x^2\mathbf{1}[-b,b](x)\ d\mu(x)\ d\mu(y)+\int_{\mathbb R}\int_{\mathbb R} y^2\mathbf{1}[-b,b](y)\ d\mu(x)\ d\mu(y)\\
&\quad -\int_{\mathbb R}\int_{\mathbb R} x^2\mathbf{1}[-b,b](y)\ d\mu(x)\ d\mu(y)-\int_{\mathbb R}\int_{\mathbb R} y^2\mathbf{1}[-b,b](x)\ d\mu(x)\ d\mu(y)\\
&=2\int_{\mathbb R} x^2\mathbf{1}[-b,b]\ d\mu-2\int_{\mathbb R}\mathbf{1}[-b,b]\ d\mu \int_{\mathbb R}x^2\ d\mu.
\end{align*}
Thus the inequality follows since $I\leq 0$.
Note. The result and its proof directly generalize to the case when $x^2$ is replaced by any even function which is non-decreasing on $[0,\infty)$ (and is integrable on bounded intervals).
