Let $D$ be a partition of a space $X$ with the quotient topology. Show that $D$ is $T_1$ if and only if the members of $D$ are closed.
We will first suppose that $D$ is $T_1$. I am then to show that the "members" of $D$ are closed. First question, what is meant by a member? $D$ consists of sets of subsets of $X$ i.e $D \subset \mathscr{P}(X)$ so am I to consider $A \in D$ or $A \subset D$?
I know that by definition of the quotient topology a subset $S \subset D$ is closed if and only if $q^{-1}(S)$ is closed in $X$. Also by definition if a space is $T_1$, then every singleton is closed. How can I apply these here? I am very confused about the wording of the problem.
The set $D$ is a partition of $X$, which means that $D$ is a set $\{D_\lambda\mid\lambda\in\Lambda\}\subset\mathcal P(X)$ such that $\lambda\ne\eta\implies D_\lambda\cap D_\eta=\emptyset$ and that $X=\bigcup_{\lambda\in\Lambda}D_\lambda$. On $D$, we can consider the quotient topology: a subset $A$ of $D$ is a set of the form $\{D_\lambda\mid\lambda\in\Lambda^\ast\}$, with $\Lambda^\ast\subset\Lambda$, and we say that $A$ is open when $\bigcup_{\lambda\in\Lambda^*}D_\lambda$ is an open subset of $X$.
Suppose that each element of $D$ is a closed set. In other words, for each $\lambda\in\Lambda$, $D_\lambda$ is a closed set. But then$$D\setminus\{D_\lambda\}=\{D_\eta\mid\eta\in\Lambda\setminus\{\lambda\}\},$$which is an open set, since its complement ($D_\lambda$) is closed. So, each singleto in $D$ is closed; in other words, $D$ is a $T_1$ space.
And if there is some $\lambda\in\Lambda$ such that $D_\lambda$ is not closed, then a similar argument shows that $\{D_\lambda\}$ is not closed and that therefore $D$ is not $T_1$.