**Claim**. *If $\|P-Q\|\leq\varepsilon\leq\frac{1}{2}$, then $|H(P)-H(Q)| \leq H(\varepsilon) + \varepsilon\log N$.*

*Proof*.
Let $\varepsilon':=\|P-Q\|$.
Let $(X,Y)$ be an optimal coupling of $P$ and $Q$, so that
\begin{align}
   \mathbb{P}(X\neq Y) = \|P-Q\| \;.
\end{align}
Using a standard construction, we can assume that $X$ and $Y$ have the particular form
\begin{align}
   X &:=
      \begin{cases}
         Z & \text{if $B=0$,} \\
         \tilde{X} & \text{if $B=1$,}
      \end{cases} &
   Y &:=
      \begin{cases}
         Z & \text{if $B=0$,} \\
         \tilde{Y} & \text{if $B=1$,}
      \end{cases}
\end{align}
where $B$, $Z$ and $(\tilde{X},\tilde{Y})$ are independent and $B\sim\text{Bern}(\varepsilon')$.

Note that
\begin{align}
   H(X|B) \leq H(X) \leq H(B) + H(X|B) \;.
\end{align}
For $H(X|B)$ we can write
\begin{align}
   H(X|B) &= \varepsilon' H(X|B=1) + (1-\varepsilon') H(X|B=0) \\
   &= \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) \;.
\end{align}
Thus,
\begin{align}
   \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) &\leq H(X)
   \leq H(B) + \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) \;,
   \tag{$\clubsuit$}
\end{align}
and similarly,
\begin{align}
   \varepsilon' H(\tilde{Y}) + (1-\varepsilon') H(Z) &\leq H(Y)
   \leq H(B) + \varepsilon' H(\tilde{Y}) + (1-\varepsilon') H(Z) \;.
   \tag{$\spadesuit$}
\end{align}

Combining ($\clubsuit$) and ($\spadesuit$) we get
\begin{align}
   |H(X)-H(Y)| &\leq
      H(B) + \varepsilon' |H(\tilde{X}) - H(\tilde{Y})| \\
   &\leq H(\varepsilon') + \varepsilon' \log N \\
   &\leq H(\varepsilon) + \varepsilon \log N \;,
\end{align}
as claimed.
QED