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

As shown by user Iosif, $M:=U/V$ has Cauchy distribution with CDF $F_R(m) := \mathbb P(R \le m)$ 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 $$ \sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon). $$

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

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 $$ \sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon). $$

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

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

Thus, noting that $\arctan$ is $1$-Lipschitz continuous, we get (with the $m_j$'s as defined in user Iosif's answer) $$ p_X(t_0,\epsilon)=\mathbb P(m_1 < M < m_2) = F_M(m_2)-F_M(m_1) = \dfrac{\arctan(m_2)-\arctan(m_2)}{\pi} \le \dfrac{m_2-m_1}{\pi}. $$

Now, fix $\alpha \in (0,1)$, and take any $t_0 \in [-\alpha,\alpha]$ and $\epsilon$ small enough that $[t_0-\epsilon,t_0+\epsilon] \subseteq [-\alpha,\alpha]$.

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 $[-\alpha,\alpha]$ with $C_\alpha = |r'(\alpha)|=1/(1-\alpha^2)^{3/2}<\infty$. 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 $$ \sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon). $$

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

As shown by user Iosif, $M:=U/V$ has Cauchy distribution with CDF $F_R(m) := \mathbb P(R \le m)$ 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 $$ \sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon). $$