= Solution
For a nonempty <open subset> $U$ of the quotient, $p^{-1}(U)$ is nonempty and <open> in the cofinite space. Therefore
$$
p^{-1}\bigl((X/\mathcal R)\setminus U\bigr)=X\setminus p^{-1}(U)
$$
is finite. <Surjectivity> implies that $(X/\mathcal R)\setminus U$ is finite as well: it is the <image> of that finite <preimage>. Hence every quotient-open set is cofinite or empty, proving the desired inclusion.
\b[Equality is not necessary.] Give $X=\mathbb Z$ its <cofinite topology> and use two equivalence classes, the even and odd integers. Neither singleton of the quotient is <open>, since its <preimage> and its <complement> are both infinite. The quotient is the two-point <indiscrete topology>, whereas the <cofinite topology> on a finite two-point set is discrete.
$$
\boxed{\mathcal T_{\mathrm{quotient}}\subsetneq\mathcal T_{\mathrm{cofinite}}\text{ can occur}.}
$$
This is the <quotient topology of a cofinite space> phenomenon, not a failure of <continuity> of the projection.
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