I'm looking for a $T_1$-space (or a $T_0$-space) and an equivalence relation ~ on $X$ such that the resulting quotient space $X/$~ is not $T_0$.
My idea: take $X=\mathbb{R}$ with the usual euclidean topology and let ~ the equivalence relation on $X$ that identifies all rationals numebers to one point $q$ and all irrationals ones to one point $i$; now if $U$ is an open not empty subset of $\mathbb{R}/\mathbb{Q}$ and $\pi:\mathbb{R}\to\mathbb{R}/\mathbb{Q}$ is the natural projection, we have that $U=\pi(\pi^{-1}(U))$; since $\pi$ is continuous, we have that $\pi^{-1}(U)$ is open, then there exists an open interval $(a,b)\subseteq\mathbb{R}$ such that $(a,b)\subseteq\pi^{-1}(U)$; hence $\pi(a,b)\subseteq U$; let $c$ e $d$ respectively a rational and an irrational numbers of $(a,b)$; so we have that $\mathbb{Q}=\pi(c)\in\pi(a,b)\subseteq U$ and $\mathbb{R}-\mathbb{Q}=\pi(d)\in\pi(a,b)\subseteq U$, then $\mathbb{Q},\mathbb{R}-\mathbb{Q}\in U$ that means $U=\mathbb{R}/\mathbb{Q}$. In conclusion we have that the topology of $\mathbb{R}/\mathbb{Q}$ is the trivial topology and that $\mathbb{R}/\mathbb{Q}$ is not $T_0$.
My questions: Do you think that my idea can work? Do you think that my example is too easy? Surely there is a more interesting one. Can anyone suggest an example satisfying the conditions of the title of my question and such that $X$ is not $T_2$? Maybe for this purpose I can take $X=\mathbb{R}$ with the upper semicontinuity topology $T=\{(-\infty,a)\mid a\in\mathbb{R}\}\cup\{\emptyset\}$ (in this case $(X,T)$ is $T_0$ but not $T_1$), what do you think?
Your example works. Here is one that starts with a $T_0$ space that is not even $T_1$.
Let $X=\{0,1,2,\ldots,n\}$ for some $n\ge 2$. For each $k\in X$ let $U_k=\{\ell\in X:\ell\le k\}$; then $\tau=\{\varnothing\}\cup\{U_k:k\in X\}$ is a $T_0$ topology on $X$ that is not $T_1$. Define the equivalence relation $\sim$ on $X$ by $k\sim\ell$ iff $k=\ell$ or $\{k,\ell\}=\{0,n\}$. Let $\pi:X\to X/\sim$ be the quotient map. Suppose that $V$ is open in $X/\sim$; then $\pi^{-1}[V]$ is open in $X$, so $0\in\pi^{-1}[V]$. But then $\{0,n\}\in V$, so $n\in\pi^{-1}[V]$. The only open subset of $X$ that contains $n$ is $X$ itself, so $\pi^{-1}[V]=X$, and therefore $V=X/\sim$. In other words, $X/\sim$ is the indiscrete space with $n$ points, which is not $T_0$.
I’m not at all sure that there are any really interesting examples of the phenomenon.