The constants in this problem do not make much sense unless $X_1$ and $X_2$ have variance $1$ so that $X_1$ and $X_2-\mu_2$ are standard normal random variables, an assumption that the OP apparently is unwilling to make since this was asked about
in the comments, and the OP did not include the assumption in the revised version of the question.
Assumption: $X_1$ and $X_2$ have variance $1$.
If $X_1$ is a standard normal random variable, then
$P\{|X_1| \geq \Phi^{-1}(1-y/2)\} = y$.
This result holds for $X_2$ as well if $\mu_2 = 0$. Thus, if
$X_1$ and $X_2$ both are independent standard normal random variables,
then
$$\begin{align}
&\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2),
|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2)\right\}
P\left\{|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= \alpha^2
\end{align}$$
while
$$\begin{align}
&\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4),
|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4)\right\}
P\left\{|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= (\alpha/2)\alpha = \alpha^2/2
\end{align}$$
In short, for the case $\mu_2 = 0$, the conjectured result holds
(with equality) for the case $\rho = 0$. Continuing to look at the
case $\mu_2 = 0$, if $X_2 = \pm X_1$ (the case when $\rho = \pm 1$),
we have
$$\begin{align}
&\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2),
|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= \alpha
\end{align}$$
while
$$\begin{align}
&\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4),
|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\
&= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4)\right\}\\
&= \alpha/2
\end{align}$$
and so once again the conjectured result holds
(with equality). What happens for other values of
$\rho \in [-1,1]$ is not immediately obvious since
the bivariate cumulative normal distribution function
must be used and the special meanings of $\Phi^{-1}$
are lost. The case $\mu_2 \neq 0$ only exacerbates the
messiness of the calculations. Simulation might be
the best option to check whether the conjectured bound
is reasonable.