The coreof the question relies on how to define for the support of a function with $x$ in a domain $\mathcal{D}$. The most common notion in DSP is the closure of the set where $f$ does not vanish, i.e. $$S=\overline{\{x\in\mathcal{D}:f(x)\ne0\}}\,,$$ hence this as a closed set.
Here, clearly, for a cardinal sine, the closure of $\mathbb{R}/ \mathbb{Z^*}$ is $\mathbb{R}$ (hence not finite).
There are general results, some even stronger (and beyond my reach), such as in
Michael Benedicks, Fourier transforms of functions supported on sets of finite Lebesgue measure, J. Math. Anal. Appl., 106 (1985), pp. 180-183. In a mundane way: if a function and its Fourier transform both have finite measure, then the function is zero almost everywhere.
Yet there are more involved notions for support, like the essential support, which depend on a measure. And you can have a (somewhat pathological) function with a support of $1$, but a $0$ essential support (like the Dirichlet function). This is not the case for continuous functions, like the triangle, the cardinal sine or its square, where both support notions coincide, see for instance Essential support vs. classical support for a continuous function. So both are good examples to that respect.
So, I suspect that the fact that $\sin(x)/x$ is not Lebesgue-integrable (and possibly the rectangular function is not continuous) played a role in your teacher's "not a good one" assertion.
PS: as evoked by Marcus Müller, note that it is somewhat funny to work in $L_2$ with projection on sine and cosine functions that are not square integrable!
Other sources :
Read More: https://epubs.siam.org/doi/10.1137/0515012