My favorite proof so far is the one from Theorem 4.5 of Yihong Wu's lecture notes, which uses the data processing inequality to reduce the problem to the binary case: if $S$ is a measurable subset of the domain, then we can consider the two random variable $\mathbf{1}_S$ under $P$ and $Q$, respectively, which has then distribution either $\mathrm{Bern}(P(S))$ or $\mathrm{Bern}(Q(S))$. First
$$
\operatorname{TV}(\mathrm{Bern}(P(S)), \mathrm{Bern}(Q(S))) = |P(S)-Q(S)| \tag{1}
$$
so we get the TV between $P$ and $Q$ by taking the supremum over all $S$. By the DPI, however,
$$
\operatorname{KL}(\mathrm{Bern}(P(S))\ \|\ \mathrm{Bern}(Q(S)))
\leq \operatorname{KL}(P\ \|\ Q) \tag{2}
$$

This shows that it suffices to prove the binary case of Pinsker's, which is just a matter of proving a simple inequality:
\begin{equation}
2(p-q)^2 \leq p\log\frac{p}{q} + (1-p)\log\frac{1-p}{1-q} \tag{3}
\end{equation}
The cases where either $p$ or $q$ is in $\{0,1\}$ are easily checked , so we can assume $p,q\in(0,1)$.
To prove (3) in this case, we introduce the function $f\colon(0,1)\to\mathbb{R}$ defined by $f(x) = p\log x + (1-p)\log(1-x)$, and observe that the RHS of (3) is exactly $f(p)-f(q)$. We then can write
$$
f(p)-f(q) = \int_{q}^p f'(x)dx = \int_{q}^p \frac{p-x}{x(1-x)}dx \geq 4\int_{q}^p (p-x)dx = 4\cdot\frac{1}{2}(p-q)^2
$$
which is (3) (in the middle, we used the fact that $x(1-x) \leq 1/4$ for $x\in(0,1)$).

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