I am trying to show that the output of the inverse Weierstrass p-function applied to some real argument is real valued. I have
$$ \alpha = \wp^{-1}(E_1 + z) $$ where $E_1,z \in \mathbb{R}$. I want to show that $\alpha$ is real-valued. Of course this isn't always true. In my case $g_2, g_3\in\mathbb{R}, \omega_1\in \mathbb{R}$ and $\omega_3\in i\mathbb{R}$. So one counter-example is $E_1 + z = e_3 \in \mathbb{R}$ and $\alpha = \omega_3$. However, I am working in a special case and all of the numerical evidence I have produced implies that in fact $\alpha$ is real valued in this case. I have $E_1 = e_1 + \sqrt{(e_1 - e_2)(e_1 - e_3)}>0$ and am considering $z\geq 0$ (so $E_1 + z > 0$ which is important as mentioned below in the comments). Of course for $z=0, \alpha = \omega_1/2$ and I think this plays a role. In my attempts to show this I have been writing
$$ \frac{\omega_1}{2} + \alpha = \wp^{-1}(E_1 + z) $$ and trying to take advantage of the information I have since it clearly is important here. It is also interesting to note that my numerical evidence also points to the exact opposite of this when $z<0$, e.g. $\alpha \in i\mathbb{R}$.
I am aware of the Whittaker and Watson exercise which shows that in this case $\wp$ takes on real values on the fundamental lattice, but this does not imply that it doesn't take on real values away from the lattice. Further I am aware that if $\omega_1/2+\alpha \in \mathbb{R}$ is a solution then so is $\omega_1/2+\alpha + 2\omega_3 \notin \mathbb{R}$ so I want an answer mod the lattice.
I use the assumptions in examples 1 and 2 from §20.32 in Whittaker's and Watson's A course of modern analysis that you seem to have followed:
Then $\omega_1$ is the real half-period, $\omega_3$ the purely imaginary half-period, and $\wp(z)$ is real on the perimeter of the rectangle with vertices $0,\omega_1,\omega_1+\omega_3,\omega_3$ except at $0$ itself where $\wp$ has a double pole. Cf. related question.
By Whittaker and Watson you know that on the real half-open line segment $z\in I=\left(0,\omega_1\right]$, $\wp(z)$ is real. Furthermore, $\wp(z)$ has no upper bound on $I$ due to its double pole at $0$ and the fact the corresponding coefficient in its Laurent series is positive. Moreover, $\wp(z)$ is continuous and reaches $\wp(\omega_1)=e_1$. Therefore, $\{\wp(z)\mid z\in I\}$ includes all real $y\geq e_1$, including your $E_1$. This, together with the symmetry $\wp(z)=\wp(−z)$, ensures the existence of two reals $\alpha$ modulo the period lattice such that $y=\wp(\alpha)$, for a given $y > e_1$. By properly considering $e_1$ as double value and counting multiplicities, this observation can be extended to $y \geq e_1$.
Now you just need to use the fact that $\wp$ takes on every complex value exactly twice over a period lattice cell, when counting multiplicities, so there cannot be more than the already found choices for $\alpha$ modulo the period lattice. As a result, there are only real $\alpha$, provided that $y \geq e_1$.
A similar argument establishes that for real $y \leq e_3$, there are two purely imaginary $\alpha$ modulo the period lattice, repeating the same value $\omega_3$ in case of $y=e_3$.
For real $y\in(e_3,e_2)$, one $\alpha$ is to be found on the line segment joining $\omega_3$ and $\omega_1+\omega_3$ because $\wp$ is real and continuous there; the other choice for $\alpha$ is given again by symmetry of $\wp$.
For $y=e_2$, we can count $\alpha=\omega_2$ twice.
For real $y\in(e_2,e_1)$, one $\alpha$ is to be found on the line segment joining $\omega_1+\omega_3$ and $\omega_1$; the other choice for $\alpha$ is given again by symmetry of $\wp$.