Define the total variation norm in the space of probability measures $\mathcal{P}(\mathbb{R}^d)$ by $$ \Vert \mu - \nu \Vert = \sup_D\lbrace | \mu(D) - \nu(D)| \rbrace. $$ I am to show that for every $\varepsilon > 0$, I find a $\delta > 0$ such that whenever $\Vert \mu - \nu \Vert < \delta$, there are measures $\tau, \mu', \nu'$ such that \begin{align*} \mu &= (1-\varepsilon)\tau + \varepsilon \mu', \\ \nu &= (1- \varepsilon)\tau + \varepsilon \nu'. \end{align*} It would suffice to show that there is a $\tau$ such that $\mu, \nu \geq (1-\varepsilon)\tau$. I have tried doing this by choosing $\tau$ as the center of a ball of diameter $\delta$ containing $\mu$ and $\nu$ so that $(1-\varepsilon)\tau$ would then drop too far from the ball, causing the inequality, but this obviously isn't working since $\tau$ can be very close to $0$ and the difference between $(1 - \varepsilon)\tau$ and $\tau$ would be insignificant compared to $\delta$.
Playing with minimums of $\mu(D), \nu(D)$ isn't working either since I can't build a $\sigma$-additive measure by combining them.
I know that the space we are in is a complete metric space.
Let $\lambda = \mu +\nu$. Let $f$ and $g$ be the densities of $\mu $ and $\nu $ w.r.t. $\lambda$ Let $\tau (A)=\int_A \min \{f,g\} d\lambda$ Then $\tau \leq \mu$ and $\tau \leq \nu$. Of course $\tau $ is not a probability measure. Can you now use that fact that $||\mu -\nu || <\delta$ to see that normalized form of $\tau $ works? [Roughly speaking $f$ and $g$ are almost the same and $\tau $ is almost a probability measure].