Closed Subgroups of $\mathbb{R}$
Hint: Show that if $F$ is not discrete, then $F$ is dense. Do this by assuming $F$ has a limit point $x_0$ (which is necessarily in $F$), and then for every $\varepsilon>0$ find some $x_\varepsilon\in F$ such that $|x_0-x_\varepsilon|<\varepsilon$. Consider translations to show that every neighborhood of radius $\varepsilon$ has an element of $F$ in it.
Thus, if $F$ isn't $\mathbb{R}$, it must be discrete. Then, use the standard technique of showing that $\inf \{x\in F:x>0\}$ generates $F$.