**Disclaimer.** *This is just to push the accepted answer a bit further and obtain an explicit upper-bound, valid for small $\epsilon$.*

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As shown by user Iosif, $M:=U/V$ has Cauchy distribution with CDF $F_M(m) := \mathbb P(M \le m)$ given by
$$
F_M(m) = \dfrac{1}{2}+\dfrac{\arctan(m)}{\pi}.
$$

Thus, if the $m_j$'s are as defined in user Iosif's answer, we ge
$$
\begin{split}
p_X(t_0,\epsilon) = \mathbb P(m_1 < M < m_2) = F_M(m_2)-F_M(m_1) &= \dfrac{\arctan(m_2)-\arctan(m_1)}{\pi}\\
& \le \dfrac{m_2-m_1}{\pi},
\end{split}
$$
where we have used the fact that $\arctan$ is nondecreasing and $1$-Lipschitz continuous.

>Now, fix $\alpha \in (0,1)$ and define $C_\alpha := 1/(1-\alpha^2)^{3/2} < \infty$.

The derivative of $r:t \mapsto -t/(1-t)^{1/2}$ is $r'(t) = -1/(1-t^2)^{3/2}$ (for $|t| < 1$), and so $r$ is $C_\alpha$-Lipschitz on the interval $[-\alpha,\alpha]$. We deduce that $m_2-m_1 \le 2C_\alpha\epsilon$.

>Thus, for any fixed $\alpha \in (0,1)$ and $\epsilon \to 0^+$, we have the following upper-bound
$$
\sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon).
$$