**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


----------


**Edit** [2018--09--17] (following *Iosif Pinelis*'s comment).  
Refining the above reasoning a little bit, we can get the better bound
\begin{align}
   |H(P)-H(Q)|\leq H(\varepsilon) + \varepsilon\log(N-1) \;.
\end{align}

Indeed, let $\Sigma$ denote the $N$-element set that is the support of $P$ and $Q$.  As before, let $\varepsilon':=\|P-Q\|$, and let us discard the trivial cases $\varepsilon'\in\{0,1\}$, so that $0<\varepsilon'<1$.

Recalling from the construction of an optimal coupling, define for $a\in\Sigma$,
\begin{align}
   R_0(a) &:= P(a)\land Q(a) &
      P_0(a) &:= P(a)-R_0(a) \\
   & &
      Q_0(a) &:= Q(a)-R_0(a) \;.
\end{align}
Observe that $R_0$, $P_0$ and $Q_0$ are non-negative functions and
\begin{align}
   \sum_{a\in\Sigma}R_0(a)=1-\varepsilon' \qquad\text{and}\qquad
      \sum_{a\in\Sigma}P_0(a)=\sum_{a\in\Sigma}Q_0(a)=\varepsilon' \;.
\end{align}
Thus, $\tilde{R}:=R_0/(1-\varepsilon')$, $\tilde{P}:=P_0/\varepsilon'$ and $\tilde{Q}:=Q_0/\varepsilon'$ are probability distributions on $\Sigma$ satisfying
\begin{align}
   P(a) &= (1-\varepsilon')\tilde{R}(a) + \varepsilon'\tilde{P} \\
   Q(a) &= (1-\varepsilon')\tilde{R}(a) + \varepsilon'\tilde{Q} \;.
\end{align}
If we now choose $Z\sim\tilde{R}$, $\tilde{X}\sim\tilde{P}$, $\tilde{Y}\sim\tilde{Q}$ and $B\sim\text{Bern}(\varepsilon')$ independently, we have a coupling as promised above.

Back to the inequality, observe that since both $P$ and $Q$ are non-negative and normalized, there necessarily exist $a,b\in\Sigma$ such that $P_0(a)=0$ and $Q_0(b)=0$.  This means that each of $\tilde{P}$ and $\tilde{Q}$ is in fact supported on a strict subset of $\Sigma$.  Hence,
\begin{align}
   |H(\tilde{X})-H(\tilde{Y})| &\leq \max\{H(\tilde{P}),H(\tilde{Q})\}
      \leq \log(N-1)
\end{align}
and the (updated) claim follows as before.

**Note**.  
As the example *H A Helfgott* gave in the comments ($N=0$, $X\sim\text{Bern}(\varepsilon)$ and $Y\sim\text{Bern}(0)$) shows, this refined bound is sharp at least when $N=2$.