Does increasing 2nd moment about "median" imply increasing variance? This is a continuation of: 
Variance of X vs Variance of a binary function of X
In that thread, I posted a partial answer, which was accepted.  However my answer has a gap.  I'm now asking for help to fix the gap.
Let $X$ be a (not necessarily continuous) random variable in $[0, 1]$, with this additional property: Suppose there exists $m \in (0, 1)$ s.t. $P(X < m) = P(X > m)$.  Define random variable $Y$ by "pushing" the probability "mass" on each side of $m$ to the limit values:


*

*If $X < m$ then $Y = 0$.

*If $X > m$ then $Y = 1$.

*If $X = m$ then $Y = m$.
Question: Is $Var(X) \le Var(Y)$?  (If true, this would complete my proof in the old thread.)
Thoughts: I'm conjecturing that $Var(X) \le Var(Y)$ should be true, based on some "moment of inertia" considerations.  By construction, the 2nd moment about $m$ has increased, i.e. $E[(Y-m)^2] \ge E[(X-m)^2]$.  If $m = E[X] = E[Y]$ then the conjecture follows at once.  The problem is that $m, E[X], E[Y]$ can be 3 different values, and I am unable to track/bound them enough to prove the conjecture.
Incidentally, there are 2 cases to this problem.  I would like an answer that covers both cases:
(1) $\forall x \in [0, 1]: P(X=x) = 0$.  In this case, $m$ always exists and is the median, and $P(X=m)=0, P(X<m) = P(X>m) = 1/2$.
(2) $\exists x \in [0,1]: P(X=x) > 0$.  In this case, $m$ might not exist (e.g. $X = \{0, 0.5, 1\}$ with probabilities $\{0.2, 0.4, 0.4\}$ respectively).  This question presupposes that $m$ exists, in which case $P(X=m) = p$ might be positive or zero.  (The $p>0$ subcase is what I need to complete my old proof.)
 A: For ease, redefine $X = X-m$ and $Y = Y-m$ (this doesn't change variances). Let $x = P(X=0)$. Then $X = 0$ with prob $x$, $X > 0$ with prob $\frac{1-x}{2}$, and $X < 0$ with prob $\frac{1-x}{2}$. We first change $X$ to be $0$ with prob $x$, $X = 1-m$ with prob $\frac{1-x}{2}$, and $X < 0$ has the same distribution as before. The change in variance is $$\left[(1-m)^2(\frac{1-x}{2})+\int_{X < 0} X^2 - \left((\frac{1-x}{2})(1-m)+\int_{X < 0} X\right)^2\right] - \left[\int_{X > 0} X^2+\int_{X < 0} X^2 - \left(\int_{X > 0} X + \int_{X < 0} X\right)^2\right].$$ Since $\int_{X < 0} X < 0$ and $\int_{X > 0} X \le (1-m)(\frac{1-x}{2})$, the term $$-2(\frac{1-x}{2})(1-m)\int_{X < 0} X + 2\left(\int_{X > 0} X\right)\int_{X < 0} X$$ is non-negative, so it suffices to show $$\int_{X > 0} X^2 - \left(\int_{X > 0} X\right)^2 \le (1-m)^2(\frac{1-x}{2})-(1-m)^2(\frac{1-x}{2})^2.$$ Note $$\int_{X > 0} X^2 - \left(\int_{X > 0} X\right)^2 \le (1-m)\int_{X > 0} X - \left(\int_{X > 0} X\right)^2.$$ The quadratic $(1-m)y-y^2$ is increasing for $y \le \frac{1-m}{2}$, so since $\int_{X > 0} X \le (1-m)(\frac{1-x}{2})$, we get the desired inequality. To go from this changed $X$ to $Y$, just do the analogous change for $X < 0$ and run through the same proof.
