Since the OP was kind enough to reference an older answer of mine, and also to alert me to the fact, I will provide some input here, of naive flavor, I guess, since mathoverflow is definitely over and out of my league. I will use $\Phi()$ for the standard normal CDF and $\phi()$ for the standard normal PDF.
In my answer in math.se I had proved that

$${\rm Var}(Y) = {\rm Var}(X)\cdot \left[1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\right],\;\; -1 <\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} \leq 0 $$

where:

$$\ln H(\mu)=\ln \big(\Phi(\beta(\mu))-\Phi(\alpha(\mu))\big)$$

$$\alpha=(a-\mu)/\sigma, \;\beta=(b-\mu)/\sigma$$.

The weak inequality comes from the fact that $H$ is log-concave (see the original post).

Calculating the second derivative, in order to prove strict inequality, we have to show that

$$\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} < 0 \implies D \equiv [-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2 < 0 $$

We note that $\Phi(\beta)-\Phi(\alpha) >0 $ always.

**CASE A : $a \leq \mu \leq b$** (including one-sided trunctaions).

Here, by the unimodality of $\phi()$, with $\mu$ being the mode, we have that

$$\phi'(\alpha) \geq 0, \phi'(\beta) \leq 0 \implies -\phi'(\alpha)+\phi'(\beta) < 0$$

since the two derivatives cannot be both equal to zero.

Then $D < 0$ always (including truncation symmetric around the mean, where, due to the fact that $\phi()$ is an even function, we have that $\phi(\alpha)=\phi(\beta)$, and the second element of $D$ will be zero. But the first element is always strictly negative).

So for this case we have proved that $\partial^2 \ln H(\mu)/\partial \mu^2 < 0$ as we wanted.

The cases left out are all the cases where $\mu$ does not belong to the (two-sided or one-sided) truncated support.

**CASE B : Truncated support $S_B\equiv (-\infty , b], \mu \notin S_B$**

Here the inequality to prove is

$$D_B \equiv \phi'(\beta)\Phi(\beta)-\phi(\beta)^2 < 0$$

Since $\phi'(\beta) = -\beta \phi(\beta)$ we wan to show

$$-\beta \phi(\beta)\Phi(\beta)-\phi(\beta)^2 < 0 \implies -\beta \Phi(\beta)-\phi(\beta) < 0 \implies \beta \Phi(\beta)+\phi(\beta) > 0$$

But this holds because

$$\beta \Phi(\beta)+\phi(\beta) = \int_{-\infty}^{\beta}\Phi(t){\rm d}t$$

and the integral is necessarily strictly greater than zero since $\Phi()$ is non-negative and non-constantly zero. So we have proved here too what we needed to prove.

**CASE C : Truncated support $S_C\equiv [a, \infty ), \mu \notin S_C$**

Here the inequality to prove is

$$-\phi'(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 \implies \alpha \phi(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 $$

and simplifying and using the symmetry properties of the two functions we have

$$ \implies (-\alpha)\Phi(-\alpha) + \phi(-\alpha) >0$$
and we are at the same situation as in Case B. So QED here too.

I am not treating the two-sided truncation cases $a < b < \mu$ and $\mu < a < b$. As with the cases I treated, I am certain there exists some more advanced and elegant way to prove what we want.