It suffices to show that there exists a sequence of arbitrarily short vectors, i.e. integer sequences $a_n, b_n$, such that
$$|a_n\sqrt{3}+b_n\sqrt{5}| \to 0$$
For this explicit example, note that an obvious choice of $a_n$, $b_n$ is
$$a_n\sqrt{3} = -b_n\sqrt{5}\iff -\frac{a_n}{b_n} = \sqrt{5/3}.$$
One can't exactly choose $a_n, b_n$ to satisfy this inequality ($\sqrt{5/3}$ is irrational).
Instead, we choose sequences $a_n/b_n$ of increasingly good rational approximations of $-\sqrt{5/3}$.
One can do this explicitly via continued fractions.
For example, wolfram alpha states the relevant continued fraction is $[1, 3, 2, 3, 2,\dots]$, meaning $1 + \frac{1}{3 + \frac{1}{2+\frac{1}{3+\dots}}}$.
Anyway, the various finite truncations of this continued fraction will be a sequence of increasingly good rational approximations to $\sqrt{5/3}$.
As the continued fraction is infinite, we can keep repeating this process, leading to $a_n, b_n$ such that $|a_n\sqrt{3}+b_n\sqrt{5}|\to 0$.
If $-\sqrt{5/3}$ were rational, its continued fraction would be finite, and the above argument would eventually fail.
Note that this means that something like the subgroup generated by $\sqrt{3}$ and $5\sqrt{3}$ is a lattice --- it is the lattice $\sqrt{3}\mathbb{Z}$.
It is important that we didn't consider the subgroup generated by $\sqrt{3}, 5\sqrt{3}$ and 1 here --- this is no longer a lattice, by (roughly) the same argument.
