quotient topology doesn't preserve separation axioms
Let $X$ be $\mathbb{R}$ in the usual topology, define an equivalence relation on $X$ by $x \sim y$ iff $x-y \in \mathbb{Q}$, let $Y$ be the set of classes, and let $q$ be the standard map that sends a point to its class. Then $Y$ in the quotient topology induced by $Y$ is uncountable and indiscrete (only $\emptyset$ and $Y$ are open), so that $X$ has all "nice" separation axioms ($T_0$ up to and including $T_6$) while $Y$, its quotient image, has none of them. I gave a proof here.
We have a similar example with $X$ and the equivalence relation where the two only classes are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, and we map to a two point set $\{0,1\}$, (say $q(x) = 0$ for $x \in \mathbb{Q}$, and $1$ otherwise) which is also indiscrete (as for $\emptyset \neq O \subseteq \{0,1\}$ if the set $q^{-1}[O]$ is open, then being non-empty it contains rationals and irrationals, so that $\{0,1\} \subseteq q[q^{-1}[O]] = O$). Both maps above are open maps even.
If $X$ is $T_2$ (Hausdorff) and not regular, and the latter is witnessed by $x \notin A$, and $A$ closed. Then indentifying $A$ to a point, gives us a quotient map onto a space that is $T_1$ but not $T_2$. This shows that closed maps can also "kill" a separation axiom. A variation is given when $X$ is not normal but $T_3$, and we identify to witnessing closed sets to points, and we get a non-$T_2$ quotient again.
Wikipedia states:
In general, quotient spaces are ill-behaved with respect to separation axioms. The separation properties of $X$ need not be inherited by $X/\!\!\sim$, and $X/\!\!\sim$ may have separation properties not shared by $X$.
The Kolmogorov quotient yields perhaps the simplest nontrivial examples of quotients $X/\!\!\sim$ having separation properties not shared by $X$. We start by defining an equivalence relation $\sim$ on $X$ wherein $x \sim y \iff$ $x$ and $y$ have the same open neighborhoods, i.e. whenever they are topologically indistinguishable. By passing to $X/\!\!\sim$, we arrive at a space where any two points are distinguishable. So when $X$ is not $T_0$, the Kolmogorov quotient on $X$ is $T_0$. On the other hand, if $X$ already is $T_0$, then $X$ and $X/\!\!\sim$ are homeomorphic.
Though unsatisfying, note also that you can simply "glue together" all of the points of any space into one, wherein we mod out by the equivalence relation $\{x \sim y\ \ \ \forall x, y \in X\}$. In this quotient, every separation property becomes vacuously true.
Just underneath the above quoted text, Wikipedia gives a hint$^\dagger$ for constructing a space $X$ and an equivalence relation $\sim$ such that we lose separation properties by passing to the quotient space:
$X/\!\!\sim$ is a $T_1$ space if and only if every equivalence class of $\sim$ is closed in $X$.
So simply construct a $T_1$ space together with a relation such that at least one of its equivalence classes is not closed in $X$. Because there is a bijective correspondence between equivalence relations on $X$ and partitions of $X$ (a corollary of the definition of an equivalence relation), you can approach this by partitioning $X$ into subsets $\{P_\alpha\}$ where at least one of these subsets is not closed. We must then lose $T_1$-ness by passing to $X/\!\!\sim$.
$^\dagger$Proof sketch of hint:
Let $f: X \rightarrow X/\!\!\sim$ be the quotient map. This result is a consequence of the following three facts, considered together:
- $U \subset X/\!\! \sim$ is closed $\iff f^{-1}(U)$ is closed in $X$
- A space is $T_1 \iff$ every singleton set is closed
- The equivalence classes in $X$ are precisely the inverse images of the singleton sets in $X/\!\!\sim$