Let $X:\Omega \to \mathbb{R}$ a random variable and define $\def\Chi{\operatorname{\raise{0.5ex}\chi}}Y = X\Chi_{\{a\leq X\leq b\}}$, where $\Chi_A$ denotes the indicator function of the set A.
Is it true that $Var(X)\geq Var(Y)$?
Assume without loss of generality $E(X)=0$ (for $E(X)=\mu$ use $X-\mu$ and $a-\mu, b-\mu$)
Note that, although $E(X)=0$, $E(Y)$ might not be $0$.
$$Y = \begin{cases}0 &\text{ if }&\qquad X<a \\ X &\text{ if }&a \leq X \leq b \\ 0 &\text{ if }& \qquad X > b \end{cases}$$
$$\begin{align} Var(Y) &= \int(Y-E(Y))^2~dP \\[1ex]&= \int_{\{X<a\}}(Y-E(Y))^2~dP+\int_{\{a\leq X \leq b\}}(Y-E(Y))^2~dP+\int_{\{X>b\}}(Y-E(Y))^2~dP \\[1ex]&= E(Y)^2P((X<a) \cup (X>b))+\int_{\{a\leq X \leq b\}}(X-E(Y))^2~dP \\[1ex]&= E(Y)^2P((X<a)\cup (X>b))+\underset{\{a\leq X \leq b\}}{\int}\left[X^2-2XE(Y)+E(Y)^2\right]~dP \\[1ex]&\leq E(Y)^2+E(X^2)-2E(Y)^2 \\[1ex]&= E(X^2)-E(Y)^2 \\[1ex]&= Var(X)-E(Y)^2 \\[1ex]&\leq Var(X) \end{align}$$
Where the first inequality follows by integrating $X^2$ over all space and not only $\{a\leq X\leq b\}$.
Is this proof correct? Even though I wrote it, it seems strange, I feel there's a mistake I'm not seeing somewhere.
Thanks in advance.
Your conjecture is certainly not true - the fact that $Y$ can now take on the value zero (when this might not be in the support of $X$) means the variance of $Y$ could be very much higher than the variance of $X$. Contrary to your assertion, your assumption that $\mathbb{E}(X)=0$ does entail an important loss of generality. Indeed, it is when $\mathbb{E}(X)$ is far from zero that there is most potential for increasing the variance by changing some of the values to zero, which is what your random variable $Y$ is doing.