In "Handbook of categorical algebra Vol 2" from Francis Borceux, the author gives a proof that $Top$ is not cartesian closed. It seems to me that this proof can be adapted to show that the category $\omega$-$QTop$ (the name is not standard) of quotient of countably based topological spaces is not cartesian closed either.

But in the paper 'Comparing cartesian closed categories of (core) compactly generated spaces' from Martín Escardó, Jimmie Lawson, Alex Simpson, it is stated that the category of quotient of countably based topological spaces, is a cartesian closed category.

So there is probably something that I don't understand well somewhere...

I apologize for the size of my question. But I think this is necessary that I reproduce here my own adaptation of the proof from Francis Borceux's book (almost everything is identical):

Let $[0,1]$ and $\mathbb{Q}$ be topological spaces with their well-known topologies (the both spaces are second countable so there are in $\omega$-$QTop$). Let us suppose that $[0,1]^\mathbb{Q}$ exists in $\omega$-$QTop$. Then the topology on $[0,1]^\mathbb{Q}$ is in particular the coarest countably-based-quotient topology making the evaluation map $ev:[0,1]^\mathbb{Q} \times \mathbb{Q} \rightarrow [0,1]$ continuous.

The proof consists in showing that the fact right above implies that $\mathbb{Q}$ is locally compact, which is a contradiction, since it is not:

Consider now the constant function $\delta:\mathbb{Q} \rightarrow [0,1]$ mapping everything to 0.

Take any point of $q$ of $\mathbb{Q}$. Since evaluation is continuous we have open neighborhood $V$ of $q$ and $U$ of $\delta$ such that $ev(U \times V) \subseteq [0,\frac{1}{2})$

We will now show that $\overline{V}$ is compact.

Take an open cover $\bigcup W_i$ of $\overline{V}$. Add $\mathbb{Q} \setminus \overline{V}$ to the open cover.

Then we create of a topology $T$ on $C(\mathbb{Q}, [0,1])$ making the evaluation map continuous: This is the topology generated by open sets of the form $$T(A,B)=\{f|\ f(\overline{A} \subseteq B)\}$$ for $A$ basic open of $\mathbb{Q}$ such that $\overline{A}$ is included in one $W_i$ and $B$ basic open of $[0,1]$. This topology is second coutable and so $(C(\mathbb{Q}, [0,1]), T)$ is in $\omega$-$QTop$.

Sine $[0,1]^\mathbb{Q}$ is structured with the coarest topology making the evaluation map continuous, there is an open $O=T(A_1,B_1) \cap \dots \cap T(A_n,B_n)$ of $T$ such that $\delta \in O \subseteq U$.

We now show that $V \subseteq \overline{A_1} \cup \dots \cup \overline{A_n}$.

Suppose not, then take $x$ such that $x \in V$ and $x \notin \overline{A_1} \cup \dots \cup \overline{A_n}$. Since $\mathbb{Q}$ is completly regular, one can find a continuous function $f$ such that $f(\overline{A_1} \cup \dots \cup \overline{A_n})=0$ and $f(x)=1$.

Since $f$ coincides with $\delta$ on each $\overline{A_i}$ we have that $f \in O \subseteq U$ but $ev(f,x) \notin [0,\frac{1}{2})$ which is a contradiction.

So $V \subseteq \overline{V} \subseteq \overline{A_1} \cup \dots \cup \overline{A_n} \subseteq W_{l_1} \cup \dots \cup W_{l_n}$.

Thus $V$ is compact. Since we can do the same for any $q \in \mathbb{Q}$ we have that $\mathbb{Q}$ is locally compact. which is a contradiction.

So $\omega$-$QTop$ is not cartesian closed...