Let $\mu$ denote the distribution of $X$ and $\nu$ the distribution of $Y$. Consider the total variation distance $d(\mu,\nu)=\max_B\mu(B)-\nu(B)$, hence $0\leqslant d(\mu,\nu)\leqslant1$ (and $d$ is probably one half of your DTV). There exists a unique maximal positive measure, usually denoted $\mu\wedge\nu$, such that $\mu-(\mu\wedge\nu)$ and $\nu-(\mu\wedge\nu)$ are mutually singular nonnegative measures. Call $s$ the total mass of $\mu\wedge\nu$, then $d(\mu,\nu)=1-s$ is the common total mass of both measures $\mu-(\mu\wedge\nu)$ and $\nu-(\mu\wedge\nu)$. Define the pair $(X',Y')$ as follows.
Let $S$ denote a Bernoulli random variable with parameter $s=\mathrm P(S=1)=1-\mathrm P(S=0)$. Let $U$, $V$ and $W$ denote independent random variables, independent on $S$, with distributions
In words, if $S=1$, $X'=Y'=W$ while, if $S=0$, $X'=U$ and $Y'=V$.
Then $\mathrm P_U$ and $\mathrm P_V$ are mutually singular hence $\mathrm P(U=V)=0$. Thus, $[X'=Y']=[S=1]$, which implies $\mathrm P(X'=Y')=s=1-d(\mu,\nu)$.
This construction is a classical application of the coupling method and is explained, if I remember correctly, in the first chapter of Lindvall's book.