Let $\pi:X \to X/\sim$ be a quotient map. Let $A \subset X$ and define $A^{*} := \pi^{-1}(\pi(A)) \subset X$. If $A^{*}$ is open or closed in X, then the subspace topology on $\pi(A)$ is the same as the quotient topology on $A^{*}/\sim$.
My ideas so far:
A set $U \in \pi(A)$ is open in the subspace topology iff $\exists \tilde{U} \subset X/\sim$
open such that $U = \pi(A) \cap \tilde{U}$. I want to show that it is open in the quotient-topology, i.e. that $\pi^{-1}(U) \subset X$ is open. But $\pi^{-1}(U) = A^{*} \cap \:\pi^{-1}(\tilde{U})$. Now if $A^{*}$ is open in X that works. But not if it is closed. And I have no idea for the other inclusion.
Any help would be appreciated.
See the proof of Theorem 22.1 in Munkres. Using the notation that you have provided, the theorem is written as follows:
In the context of your question, the subspace $A^{*}$ is clearly saturated since it is the inverse image of a subset of the codomain $X /\sim$. Moreover, $\pi^*(A^*) = \pi(\pi^{-1}(\pi(A)) = \pi(A)$, so the restriction $\pi^*$ of $\pi$ to $A^*$ can in fact be written as $\pi^* : A^* \rightarrow \pi(A)$. Thus, the theorem tells you that $\pi^*$ is a quotient map; in particular, a subset $V \subseteq \pi(A)$ is open if and only if it is open in ${\pi^*}^{-1}(V)$ if and only if it is open in $\pi^*(A^*)$. By definition, this implies equality of the subspace topology on $\pi(A)$ and the quotient topology on $\pi^*(A^*)$.