Proof
Let $\Omega$ be finite, and let $P,Q$ be probability distributions on $\Omega$ with $|P - Q|_1 \leq \frac{1}{2}$. Denote $|P-Q|1$ by $\alpha$. Let
$H(P) = -\sum{x\in \Omega}p(x)\log p(x).$
We prove
$$
|H(P) - H(Q)| \leq \alpha \log!\left(\frac{|\Omega|}{\alpha}\right).
$$
Let $\alpha = |P - Q|1$.
Let $S = {x \in \Omega : p(x) \geq \frac{\alpha}{|\Omega|}}$.
Then $\sum{x\in S^c}p(x) \leq |S^c|\cdot \frac{\alpha}{|\Omega|} \leq \alpha$.
Hence, for all $x \in S$, $\log!\left(\frac{1}{p(x)}\right) \leq \log!\left(\frac{|\Omega|}{\alpha}\right)$.
Assume w.l.o.g. that $H(P)\geq H(Q)$. Then
$$
H(P)-H(Q) = \sum_{x\in \Omega}\left[p(x)\log\frac{1}{p(x)} - q(x)\log\frac{1}{q(x)}\right].
$$
Split the sum over $S$ and $S^c$:
$$
\sum_{x\in \Omega}p(x)\log\frac{1}{p(x)} = \sum_{x\in S}p(x)\log\frac{1}{p(x)} + \sum_{x\in S^c}p(x)\log\frac{1}{p(x)}.
$$
By line 4, for $x \in S$, $\log\frac{1}{p(x)} \leq \log\frac{|\Omega|}{\alpha}$. For $x \in S^c$, $p(x)\leq \frac{\alpha}{|\Omega|}$. Thus
$$
\sum_{x\in S^c} p(x)\log\frac{1}{p(x)} \leq \left(\frac{\alpha}{|\Omega|}\cdot|S^c|\right)\log\frac{|\Omega|}{\alpha}
\leq \alpha\log\frac{|\Omega|}{\alpha}.
$$
Let $\Delta_x = p(x)-q(x)$. Then $\sum_{x\in \Omega}\Delta_x=0$ and $\sum_{x\in \Omega}|\Delta_x|=\alpha$.
Rewrite
$$
p(x)\log\frac{1}{p(x)} - q(x)\log\frac{1}{q(x)}
= \Delta_x\log\frac{1}{p(x)} + q(x)\left[\log\frac{1}{p(x)} - \log\frac{1}{q(x)}\right].
$$
For $x$ with $\Delta_x>0$, we have $p(x)\geq q(x)$. Then $\log\frac{1}{p(x)} - \log\frac{1}{q(x)} = \log!\left(\frac{q(x)}{p(x)}\right)\leq 0$.
Thus, on those $x$ with $\Delta_x>0$, the second term in line 9 is $\leq 0$. Hence
$$
p(x)\log\frac{1}{p(x)} - q(x)\log\frac{1}{q(x)}
\leq \Delta_x\log\frac{1}{p(x)}
\leq \Delta_x\log\frac{|\Omega|}{\alpha}.
$$
Summing over all $x$ with $\Delta_x>0$ yields
$$
\sum_{\substack{x:,\Delta_x>0}}\left[p(x)\log\frac{1}{p(x)} - q(x)\log\frac{1}{q(x)}\right]
\leq \left(\sum_{\Delta_x>0}\Delta_x\right)\log\frac{|\Omega|}{\alpha}.
$$
Since $\sum_{\Delta_x>0}\Delta_x \leq \alpha$, the sum in line 12 is $\leq \alpha\log!\left(\frac{|\Omega|}{\alpha}\right)$.
A symmetric argument applies to those $x$ with $\Delta_x<0$.
Combining the bounds for $x\in S\cup S^c$ and for the cases $\Delta_x>0$ or $\Delta_x<0$, we get
$$
H(P)-H(Q)
\leq \alpha\log!\left(\frac{|\Omega|}{\alpha}\right).
$$
Reversing the roles of $P$ and $Q$ shows $\left|H(P)-H(Q)\right|\leq \alpha\log!\left(\frac{|\Omega|}{\alpha}\right)$.
$\square$
