Example of discrete subgroup of $\mathbb{R}^2$ where image of $X$ under 1st projection isn't discrete subset of $\mathbb{R}$? What is an example of a discrete subgroup $X$ of $(\mathbb{R}^2, +)$ where the image of $X$ under the 1st projection $\mathbb{R}^2 \to \mathbb{R}$ isn't a discrete subset of $\mathbb{R}$?

How about $G=\langle(1,1), (\pi, 1)\rangle?$ Then $0$ should be accumulation point of the projection.
1) $G$ is discrete:
Consider $x=(a+b \pi, a+b) \in G$ arbitrary element. Suppose that $y=(a'+b' \pi, a'+b') \in G$ is another element that is sufficiently close to $x$, say $|x-y|<\frac{1}{2}$. Then we necessarily have that $a+b=a'+b'$, since otherwise $|x-y|\geq 1$ ($a+b, a'+b'$ are integers). 
Now 
$x-y=((a-a')+(b-b')\pi, 0)$, so we have 
$$|x-y|=|(a-a')+(b-b')\pi| < \frac{1}{2}.$$
But $a+b=a'+b'$ is equivalantly $(a-a')=-(b-b')$, so we have 
$$x-y=(k-k\pi, 0)=(k(1-\pi), 0)$$
for some $k\; (=a-a')$.
It follows that $|k|(\pi-1)=|x-y|<\frac{1}{2},$
which can happen only if $k=0$ ($k$ is an integer). Thus, $a=a', b=b'$ and so $x=y$.
2) $\mathrm{proj}_1(G)$ is not discrete:
It is enough to show that $\langle 1, \pi \rangle \subseteq \mathbb{R}$ is not discrete. 
By Dirichlet's approximation theorem, There are infinitely many pairs $(p, q)$ of integers such that 
$$\left|\pi-\frac{p}{q}\right|<\frac{1}{q^2}\;.$$
That is, there are infintely many positive integers $q$ (check) such that for a suitable integer $p$ (depending on $q$) we have 
$$-\frac{1}{q^2}<\pi-\frac{p}{q}<\frac{1}{q^2},$$ hence
$$-\frac{1}{q}<q\pi-p<\frac{1}{q}.$$
This shows that $0$ is an accumulation point of $\langle 1, \pi \rangle$.
EDIT: You may notice that the example is unnecessarily complicated: Taking $\langle(1,1),(\pi, 0) \rangle$ would certainly simplify step $1$.
