= Solution
A <coupling of probability distributions> is a <joint probability distribution> of <random variables> $(X,Y)$ with the specified <marginal distributions>: $\mathbb P(X\in A)=\mu(A)$ and $\mathbb P(Y\in A)=\nu(A)$. No <independence> is required.
For every <set> $A$, the difference of the two indicator functions vanishes when $X=Y$. Consequently
$$
|\mu(A)-\nu(A)|=\left|\mathbb E\bigl(\mathbf1_A(X)-\mathbf1_A(Y)\bigr)\right|\leq\mathbb P(X\ne Y).
$$
Maximizing over $A$ gives the <coupling inequality for total variation>:
$$
\boxed{\|\mu-\nu\|_{\mathrm{TV}}\leq\mathbb P(X\ne Y).}
$$
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