= Solution
For the surjection $p$ to the set of equivalence classes, define a subset $V$ of the quotient to be open exactly when $p^{-1}(V)$ is open in $X$. Inverse images preserve the empty and full sets, arbitrary unions and finite intersections. Applying the topology axioms in $X$ proves this is a topology on the quotient, and it makes $p$ continuous.
The map need not be open. On $X=\mathbb R$, identify zero with one and leave every other point in its own class. The set $U=(-1/4,1/4)$ is open, but its <saturation> is
$$
p^{-1}(p(U))=U\cup\{1\},
$$
which is not open in $\mathbb R$. Thus $p(U)$ is not open in the <quotient topology>. <Continuity> of a <quotient map> does not imply openness.
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