It will suffice to show that there is an absolute constant $c_1>0$
such that
$$
|\omega|\geq c_1, \ \text{for any } \omega\in L\setminus
\lbrace 0 \rbrace \tag{1}
$$
Indeed, once you have (1), it follows that $|\omega-\omega'| \geq c_1$
for any distinct $\omega,\omega’$ in $L$. If we put
$A_n=\lbrace z \in {\mathbb C} \ | \ n \leq |z| \leq n+1\rbrace$ and
$B_n=A_n \cap L$, then all the disks $D(b,\frac{c_1}{3}) (b\in B_n)$ are
disjoint and contained in $A’_n=\lbrace z \in {\mathbb C} \ | \
n-\frac{c_1}{3} \leq |z| \leq n+1+\frac{c_1}{3}\rbrace$, so the total
area they cover cannot exceed the area of $A’_n$, which is
$\pi(1+\frac{2c_1}{3})(n+2)$. So the number of lattice points in $A_n$
is at most $\frac{9}{c_1^2}(1+\frac{2c_1}{3})(n+2)$.
Let us now show (1). Write $\omega_1=x_1+iy_1,\omega_2=x_2+iy_2$ where
$x_1,y_1,x_2,y_2$ are real. The hypothesis $\frac{\omega_2}{\omega_1}\not\in{\mathbb R}$
yields $d=x_1y_2-x_2y_1 \neq 0$. Put
$$
c_1={\sf min}\Bigg(\frac{|d|}{2(|x_2|+|y_2|)},\frac{|d|}{2(|x_1|+|y_1|)}\Bigg)>0 \tag{2}
$$
Let $\omega\in L$, so
that $\omega=n\omega_1+m\omega_2$ with $n,m\in\mathbb Z$.
If $|\omega|<c_1$, then $|{\sf Re}(\omega)| \leq c_1$ and
$|{\sf Im}(\omega)| \leq c_1$, so that
$$\big|x_2{\sf Re}(\omega)-y_2{\sf Im}(\omega)\big| \leq c_1 (|x_2|+|y_2|),\ \
\big|x_1{\sf Re}(\omega)-y_1{\sf Im}(\omega)\big| \leq c_1 (|x_1|+|y_1|) \tag{3}$$
Since ${\sf Re}(\omega)=nx_1+mx_2$ and ${\sf Im}(\omega)=ny_1+my_2$, it follows
from (3) that $|n|\leq \frac{1}{2}, |m|\leq\frac{1}{2}$, hence $\omega=0$. This finishes
the proof.
