Lower-bound on zero-crossing probability of the nonstationary gaussian process $X(t) = tU+(1-t^2)^{1/2}V$, with $(U,V) \sim N(0,I_2)$ Let $(X(t))_{t \in [-1,1]}$ be a centered non-stationary smooth gaussian process with covariation function $\rho(t,s) = \mathbb E[X(t)X(s)]$. For $t_0 \in (-1,1)$ and $\epsilon \in (-1-t_0,1-t_0)$, define
$$
p_X(t_0,\epsilon) : = \mathbb P(X(t) = 0\,\text{ for some } t \in [t_0-\epsilon,t_0+\epsilon])
$$

Question. What is a good upper-bound for $p_X(t_0,\epsilon)$ which is valid for small $\epsilon$ (i.e for $\epsilon \to 0^+$) ?

A concrete example. The GP I have in mind is $X(t) := tU + (1-t^2)^{1/2}V$, where $(U,V) \sim N(0,I_2)$, for which the covariation function is
$$
\rho(t,s) = ts +(1-t^2)^{1/2}(1-s^2)^{1/2}.
$$
 A: $\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example,
$$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$
where
$$m_1:=\min_{t\in[t_0-\ep,t_0+\ep]}r(t)
=r(t_0+\ep),\quad 
m_2:=\max_{t\in[t_0-\ep,t_0+\ep]}r(t)
=r(t_0-\ep),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$
this follows because $r$ is a continuous decreasing function on the interval $(-1,1)$.
Letting now
$$\th_j:=\arctan m_j$$
and using the rotational symmetry of the distribution of $(U,V)$, we get
$$p_X(t_0,\ep)=\frac{\th_2-\th_1}{\pi}.
$$
A: 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
$$
\sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon).
$$

