Definition: Suppose $X, \tau$ is a topological space and $R$ is an equivalence relation on $X$. Let $X/R$ denote the set of $R$-equivalence classes. Define the function $f$ from $X$ to $X/R$ by $f(x) = \hat x$ where $x$ is any element of $X$ and $\hat x$ is the $R$-equivalence class of $x$. Then $f$ is called the identification mapping from $X$ to $X/R$. Define a subset $U$ of $X/R$ to be open if $f^{-1}(U)$ is open in $X$. The topology thus obtained on $X/R$ is called the identification topology on $X/R$.
So, I decided to try $X = \{1,2,3\}, \tau = \{\emptyset, \{1\}, X\}$ where the equivalence classes will be $\{1,3\}, \{2\}$. I suppose now that this space is $\{\emptyset, \{1,3\}, \{2\}, \{1,2,3\}\}$. My problem is figuring out the topology on this. I imagine I could $\{1\} \to \{1,3\}$, so my topology on this should be $\{\emptyset, \{1,3\}, \{1,2,3\}\}$.
Is this correct?
Your new space is the space of classes so the space is simply $X/R = \{\{1,3\},\{2\}\}$. Now $\emptyset$ and the whole space must be open so we need to decide the openness of the singleton sets $\{\{1,3\}\}$ and $\{\{2\}\}$. The first has preimage $\{1,3\}$ which is not open and the second has preimage $\{2\}$ which is not open either. So the topology is $\{\emptyset, X/R\}$ aka the trivial topology.
This sort of thing gets less confusing if we use better notation, i.e. $[1]$ for the class containing 1!