Design of Reed-Solomon Code where $n < q-1$
You have the right idea but are missing some of the fine details.  A cyclic Reed-Solomon code of length $n$, where $n$ is a proper divisor of $q-1$, is a BCH code and the key property is that if a sequence of $2t$ consecutive $\mathbf{n}$-th roots of unity are roots of the generator polynomial, then the minimum distance is at least $2t+1$. Here, $n = 3$ and $\alpha^5$, $\alpha^{10}$ and $\alpha^{15}=1$ are the three cube roots of unity. So, choose two of them to be roots of $g(x)$ (it doesn't matter which ones because you will always get two consecutive roots in this simple case!) and you are done. If you choose $\beta = \alpha^5$ and $\beta^2 = \alpha^{10}$ as the roots of $g(x)$, then $g(x) = x^2+x+1$, and so all the codewords on this $[3,1,d=3]$ Reed-Solomon code are of the form $\gamma +\gamma x + \gamma x^2, \gamma \in$ GF$(2^4)$, or equivalently $[\gamma, \gamma, \gamma]$,   that is, we have a repetition code!

It is also the case here that the cube roots of unity are elements of the subfield GF$(2^2)$ but this is not the case in general. Take $n=5$ which is also a divisor of $q-1=15$ in which case $\alpha^3, \alpha^6, \alpha^9, \alpha^{12}$ and $\alpha^{15}=1$ are the five $5$-th roots of unity which all belong to GF$(2^4)$.