Is a Lattice always uniformly discrete in $\mathbb{C}$? The following definition of discreteness and uniform discreteness comes from Wikipedia.
A subset $S$ of a metric space $(Y,d)$ is said to be discrete in $Y$, if for all $x\in S$, there exists some $\delta=\delta(x)>0$ such that $d(x,y)>\delta$ for all $y\in S\setminus\{x\}.$
A subset $S$ of a metric space $(Y,d)$ is said to be uniformly discrete in $S$, if there exists a $\delta>0$ such that for any $x,y\in S$, either $x=y$ or $d(x,y)>\delta$.

Now, Rick Miranda's Algebraic Curves and Riemann Surfaces has the following construction.
Consider the complex plane $\mathbb{C}$, and let $w_{1}, w_{2}\in\mathbb{C}$ be linearly independent over $\mathbb{R}$, and we fix this pair $w_{1}, w_{2}$. Define the Lattice by $$L:=\mathbb{Z}w_{1}+\mathbb{Z}w_{2}=\{a_{1}w_{1}+a_{2}w_{2}:a_{1},a_{2}\in\mathbb{Z}\}.$$
Then, the books said:

The lattice $L$ is a discrete subset of $\mathbb{C}$, so there exists $\epsilon>0$ such that $|w|>2\epsilon$ for every non-zero $w\in L$. Fix such an $\epsilon$, and.....(not relevant stuff starts).

Given the context of his sentence, it seems that he is claiming that $L$ is in fact uniformly discrete in $\mathbb{C}$. (Otherwise he cannot fix such $\epsilon$, without fixing the choice of point from the Lattice, and the later proof seems assuming this $\epsilon$ works for everyone)
Is this true? I looked it up online, but did not find any convincing reference.

I tried to prove it. So, we write $w_{1}=x_{1}+iy_{1}$ and $w_{2}=x_{2}+iy_{2}$, and we let $x,y\in L$ so that $x=w_{1}a_{1}+w_{2}a_{2}$ and $y=w_{1}b_{1}+w_{2}b_{2}$, where $a_{1},a_{2},b_{1},b_{2}\in\mathbb{Z}$.
Then, the (complex norm) distance between $x$ and $y$ is
\begin{align*}
|x-y|^{2}&=|a_{1}w_{1}+a_{2}w_{2}-b_{1}w_{1}-b_{2}w_{2}|^{2}\\
&=|a_{1}x_{1}+ia_{1}y_{1}+a_{2}x_{2}+ia_{2}y_{2}-b_{1}x_{1}-ib_{1}y_{1}-b_{2}x_{2}-ib_{2}y_{2}|^{2}\\
&=(a_{1}x_{1}+a_{2}x_{2}-b_{1}x_{1}-b_{2}x_{2})^{2}+(a_{1}y_{1}+a_{2}y_{2}-b_{1}y_{1}-b_{2}y_{2})^{2}\\
&=\Bigg[x_{1}(a_{1}-b_{1})+x_{2}(a_{2}-b_{2})\Bigg]^{2}+\Bigg[y_{1}(a_{1}-b_{1})+y_{2}(a_{2}-b_{2})\Bigg]^{2}\\
&=\Bigg[x_{1}^{2}(a_{1}-b_{1})^{2}+2x_{1}x_{2}(a_{1}-b_{1})(a_{2}-b_{2})+x_{2}^{2}(a_{2}-b_{2})^{2}\Bigg]\\
&\ \ \ \ \ \ \ \ \ +\Bigg[y_{1}^{2}(a_{1}-b_{1})^{2}+2y_{1}y_{2}(a_{1}-b_{1})(a_{2}-b_{2})+y_{2}^{2}(a_{2}-b_{2})^{2}\Bigg]\\
&\geq 2x_{1}x_{2}(a_{1}-b_{1})(a_{2}-b_{2})+2y_{1}y_{2}(a_{1}-b_{1})(a_{2}-b_{2})\\
&=2(a_{1}-b_{1})(a_{2}-b_{2})(y_{1}y_{2}+x_{1}x_{2}).
\end{align*}
What can I do further  I get a lower bound of this expression that does not depend on $a_{1},a_{2}$ and $b_{1},b_{2}$? (so does not depend on $x$ and $y$).
For now, I have not utilized the fact that $a_{i},b_{i}\in\mathbb{Z}$. Well, if we have a perfect scenario that $a_{1}>b_{1}\geq 0$ and $a_{2}>b_{2}\geq 0$, then  $$2(a_{1}-b_{1})(a_{2}-b_{2})(y_{1}y_{2}+x_{1}x_{2})>2\cdot 1\cdot 1(y_{1}y_{2}+x_{1}x_{2}),$$ is the desired bound.
But I am not sure what to do if we are not in this kind of scenario (so I guess this gives us some sort of a counter-example?)
Thank you!
Edit 1:
Okay, I think Rick Miranda actually means discrete. I will explain in the answer I will post on my own. Also, I will also point a proof of why $L$ is discrete, as my above proof cannot be used to prove this either.
 A: If we find some $d > 0$ such that
$$
|w| \ge  d \text{ for all non-zero $w$ in } L = \mathbb{Z}w_{1}+\mathbb{Z}w_{2}
$$
then
$$
 |w - \tilde w| \ge d > \frac d2 
$$
for all $w \ne \tilde w$ in $L$, i.e. the lattice is uniformly discrete.
In order to find a lower bound $d$ for the distance of lattice points to the origin, one can proceed as follows:  Let $w = kw_1 + l w_2$ with $(k, l) \in \Bbb Z^2$, $(k, l) \ne (0,0)$. If $l = 0$ then $|k| \ge 1$ and
$$
|w| = |k| \cdot |w_1| \ge |w_1| \, .
$$
If $l \ne 0$ then $|l| \ge 1$ and
$$
|w| = |w_1| \cdot \left| k + l \frac{w_2}{w_1}\right|
\ge |w_1| \cdot \left| \operatorname{Im} \left( k + l \frac{w_2}{w_1}\right) \right|\\
= |w_1| \cdot |l| \cdot \left|\operatorname{Im}\left(\frac{w_2}{w_1}\right)\right|
\ge |w_1| \cdot \left|\operatorname{Im}\left(\frac{w_2}{w_1}\right) \right| \, .
$$
The right-hand side is strictly positive because $w_1, w_2$ are linearly independent over $\Bbb R$.
This shows that for all non-zero $w$ in $L$
$$
|w| \ge d = |w_1| \cdot \min \left(1,  \left|\operatorname{Im}\left(\frac{w_2}{w_1}\right)\right| \right)> 0 \, .
$$
A: Let me first quote the whole paragraph of what Rick Miranda wrote:

The Lattice $L$ is a discrete subset $\mathbb{C}$, so there is an $\epsilon>0$ such that $|w|>2\epsilon$ for every nonzero $w\in L$. Fix such an $\epsilon$ and fix a point $z_{0}\in \mathbb{C}$. Consider the open disk $D=D(z_{0},\epsilon)$. This choice of $\epsilon$ insures that no two points of $D$ can differ by an element of $L$.

I firstly thought that he was saying something of uniformity. But I think that he basically means this:
The Lattice $L$ is a discrete subset $\mathbb{C}$. Therefore, by definition, for any $x\in L$, there exists $\delta=\delta(x)>0$ such that $|x-y|>\delta$ for all $y\in L\setminus\{x\}$. Set $\epsilon:=\frac{\delta(0)}{2}$. Then, $|y|=|0-y|>2\epsilon$ holds for all $y\in L\setminus\{0\}$. Fix a point $z_{0}\in\mathbb{C}$, and and consider the open disc $D(z_{0},\epsilon)$.

So, how to show that $L$ is discrete? In the comment, Thorgott has pointed out one way, and I will give a more geometric proof.
Remember that when consider the subset $S$ as a topological space (with the subspace topology) itself, it is discrete if and only if all of its singletons are open, which happens if and only if none of its singleton contain any accumulation points.
Hence, we will show that any singleton of $L$ is open in the subspace topology. In other words, for any $z_{0}\in L$, $$\{z_{0}\}=L\cap U\ \ \text{for some open set}\ \ U\subset\mathbb{C}.$$
The idea is to find a disk $U:=D(z_{0},r)$ centered at $z_{0}$ with a sufficiently small radius $r$, then the intersection of $U$ and $S$ will only contain $z_{0}$. Note that to ensure that no $y\in L\setminus\{z_{0}\}$ belongs to $L\cap D(z_{0},r)$, we only need to set $r$ to be the minimum distance between $z_{0}$ and other points in the lattice. However if $z_{1}$ is another a point in the lattice, then $z_{1}-z_{0}$ is also a lattice. Hence, to determine the minimum distance between $z_{0}$ and other points, it is equivalent to determine the minimum distance between $0$ and other points in this lattice.
But this is easy, the point that is closest  to the origin is one of the initial point, i.e. is either $w_{1}$ or $w_{2}$. So, the minimum distance is $\min\{|w_{1}|,|w_{2}|\}$.
Hence, simply choosing $r=\min\{|w_{1}|,|w_{2}|\}$, and setting $U:=D(z_{0},r)$ finishes the proof.

Please feel free to post your proof of either discreteness or uniform discreteness, or counter-example, or comment. Thanks for everyone's help!!!
