Late answer, but here it is why.
If $g_2, g_3 \in \mathbb{R}$ and $g_2^3-27g_3^2>0$ it can bee seen that the half periods are of the form $\omega_1=\alpha$ and $\omega_2=i \beta$ with $\alpha, \beta \in \mathbb{R}$.
This essentially does the trick since it is easily seen by a direct computation that if $\wp(z)=\wp(z|\alpha, i \beta)$ then
\begin{equation}
\overline{\wp(z)}=\wp\left(\overline{z}\right) \tag{1}
\end{equation}
Now take $z \in \{ iy, \ \alpha+iy, \ x, \ x+i \beta : x,y \in \mathbb{R}\}$.
i): If $z=iy$ with $y \in \mathbb{R}$, by (1) it follows that $\overline{\wp(z)}=\overline{\wp(iy)}=\wp\left(\overline{iy}\right)=\wp(-iy)$, but since $\wp$ is even then $\wp(-iy)= \wp(iy)$, thus indeed
$$
\overline{\wp(z)}=\wp(iy)=\wp(z), \ \text{hence} \ \wp(z) \in \mathbb{R}
$$
ii): If $z=\alpha + iy$ with $y \in \mathbb{R}$, again by (1), $\overline{\wp(z)}=\overline{\wp(\alpha+iy)}=\wp\left(\overline{\alpha + iy}\right)=\wp(\alpha-iy)$, but $2\alpha$ is a period of $\wp$, that gives $\wp(\alpha-iy)= \wp(\alpha-iy-2\alpha)=\wp(-(\alpha+iy))$, again being $\wp$ even we have $\wp(-(\alpha+iy))=\wp(\alpha+iy)$, therefore
$$
\overline{\wp(z)}=\wp(\alpha+iy)=\wp(z), \ \text{hence } \ \wp(z) \in \mathbb{R}
$$
iii): If $z=x$ with $x \in \mathbb{R}$, (1) gives that $\overline{\wp(z)}=\overline{\wp(x)}=\wp(\overline{x})=\wp(x)=\wp(z)$, then of course $\wp(z) \in \mathbb{R}$
iv): Finally when $z=x + i\beta$ with $x \in \mathbb{R}$, by (1), $\overline{\wp(z)}=\overline{\wp(x+i\beta)}=\wp\left(\overline{x + i\beta}\right)=\wp(x-i\beta)$, since $2i\beta$ is a period of $\wp$ then $\wp(x-i\beta)= \wp(x-i\beta+2i\beta)=\wp(x+i\beta)$, therefore again
$$
\overline{\wp(z)}=\wp(x+i\beta)=\wp(z), \ \text{hence } \ \wp(z) \in \mathbb{R}
$$
Thus indeed $\wp$ takes real values in all the parallelograms generated by the half periods $\omega_1$ and $\omega_2$, as wanted.