Subgroup of $\mathbb{R}$ either dense or has a least positive element?
Let $G$ be an additive subgroup of $\mathbb{R}$. Suppose that there does not exists the least positive element, i.e. $\inf\{|x|:x\in G-\{0\}\}=0$. Then we can prove that $G$ is dense in $\mathbb{R}$ as follows: suppose that $y\in\mathbb{R}$, for any $\epsilon>0$, there exist $x\in G$ such that $|x|<\epsilon$. We can assume that $x>0$, otherwise, we can take $-x$ which belongs to $G$ since $G$ is an additive group. Then there exists an integer $n$ such that $$nx\leq y<(n+1)x,$$ which implies that $$|y-nx|<x<\epsilon,$$ where $nx\in G$ since $G$ is an additive group. This shows that $G$ is dense in $\mathbb{R}$.
Let $a=\inf\{g\in G; g>0\}$. (Note that this set is non-empty and thus $a\ge 0$, $a<\infty$.)
Suppose that $a\notin G$.
This implies that for each $\varepsilon>0$ there is $g\in G$ such that $a<g<a+\varepsilon$. The same argument gives the existence of $g'$ such that $a<g'<g<a+\varepsilon$. Thus we found $h_\varepsilon=g-g'$ which belongs to $G$ and fulfills $0<h_\varepsilon<\varepsilon$.
Now it is relatively easy to see that $\bigcup\limits_{\varepsilon>0}\{z\cdot h_\varepsilon; z\in\mathbb Z\}$ is dense in $\mathbb R$ and it is subset of $G$.
(If you choose some interval $(a-\varepsilon,a+\varepsilon)$, which has length $2\varepsilon$, then it must contain some element of the form $z\cdot h_\varepsilon$, since the distance between two neighboring elements of this form is less than $\varepsilon$.)
An analogous claim (or at least one of them - there are other possible generalizations) in higher dimensions would be:
An additive subgroup of $\mathbb R^n$ is discrete if and only if it is generated by linearly independent vectors $a_1,\dots,a_m\in\mathbb R^n$, $m\le n$.
In the case $n=1$ we get only one generator.
Proof of this claim can be found e.g. in Stewart, Tall: Algebraic Number Theory and Fermat's Last Theorem 6.1 (in Section 6.1 Lattices) and probably in many other texts dealing with lattices in $\mathbb R^n$.
I once used this fact to solve a problem in the American Mathematical Monthly. Here's the proof I gave, which is rather similar to Paul's.
We first show that if $G$ has a limit point, then it is dense. Fix $\epsilon > 0$ and let $y$ be a limit point of $G$. Choose $x_1, x_2 \in G$ with $0 < |{x_1-y}| < \epsilon/2$, and $|{x_2 - y}| < |{x_1 - y}|$. Then $x := x_1 - x_2 \in G$ with $0 < |{x}| < \epsilon$. $G$ now contains all integer multiples of $x$, so every interval of length at least $\epsilon$ contains an element of $G$. $\epsilon$ was arbitrary so $G$ is dense.
As the contrapositive of this, if $G$ is not dense then it has no limit points. In this case, let $a := \inf \{ x \in G, x > 0\}$. (This set is nonempty since $G$ is nontrivial and hence contains at least one positive number.) Since neither $0$ nor $a$ are limit points of $G$, $a > 0$ and $a \in G$. By construction, $a$ is the least positive element of $G$. (Moreover, $G$ is generated by $a$.)
For those curious about the Monthly problem, it's #11345, volume 115 number 2, page 166, February 2008:
11345. Proposed by Roger Cuculière, France. Find all nondecreasing functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x+f(y)) = f(f(x)) + f(y)$ for all real $x$ and $y$.