What is wrong with my proof? For $x,y\in[0,1]$ let $x\sim y$ iff $x=y$ or $x,y\in\Bbb Q$, and let $X=[0,1]/\sim$; $X$ is the result of identifying $\Bbb Q\cap[0,1]$ to a point. 
I wrote a proof that the topology on $[0,1]/\sim$ is the trivial topology, can you tell me please where is my mistake? Thank you for correcting me.
Let $\pi:[0,1]\to [0,1]/\sim$. Let $O\subseteq [0,1]/\sim$ be an open set, let $[x]\in O$. Then $\pi^{-1}O$ is open so there is $\varepsilon >0$ so that $B(x,\varepsilon)\subseteq \pi^{-1}O$. Then there is a rational $q$ so that $d(x,q)<\varepsilon / 3$ therefore $[q]\in \pi B(x,\varepsilon)\subseteq O$.
Let $\delta>0$ be such that $\pi B(q,\delta)\subseteq \pi B(x,\varepsilon /3)\subseteq O$. Let $y$ be a point in $[0,1]$. Then there is a rational $q$ with $d(y,q)<\delta$. Therefore $[y]\in \pi B(q,\delta)\subseteq \pi B(x,\varepsilon)\subseteq O$. Therefore $O=[0,1]/\sim$ is the whole space if $O\neq \varnothing$. 
 A: (For some reason I never saw your comments on my answer to the earlier question; I’m not sure what happened. Just as well, maybe: I’ve more room to answer here.)
The first half of your argument is correct: every non-empty open subset of $[0,1]/\sim$ contains $[q]$. The second half is not. Let $a$ be any irrational in $[0,1]$; then $\pi^{-1}\big[X\setminus\{[a]\}\big]=[0,1]\setminus\{a\}$ is an open set in $[0,1]$, so $X\setminus\{[a]\}$ is open in $X$, which therefore does not bear the indiscrete topology. 
The same argument can be make for any closed subset $K$ of $[0,1]\setminus\Bbb Q$: $$\pi^{-1}\big[X\setminus\pi[K]\big]=[0,1]\setminus K\;,$$ which is open in $[0,1]$, so $X\setminus\pi[K]$ is open in $X$.
The specific point of error in your argument is here when you say

Let $\delta>0$ be such that $\pi B(q,\delta)\subseteq \pi B(x,\varepsilon /3)\subseteq O$.

For any fixed $q\in[0,1]\cap\Bbb Q$ such a $\delta$ can be found, but $\delta$ depends on $q$: different values of $q$ will in general require different values of $\delta$. In the case of the open set $X\setminus\{[a]\}$ for some irrational $a$, for instance, rationals close to $a$ will require smaller values of $\delta$ than those further away from $a$.
A: I believe you made a mistake when you claimed that $\pi^{-1}O$ was an open set. In general, this will not be true in $[0, 1]$; what is true is that there is some open $A\subseteq[0, 1]$ such that $\pi(A)=O$.
