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Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap\mathcal W$ of a countable family $\mathcal W\subseteq\tau'$. Taking into account that each set $U'\in\tau'\setminus\tau$ is of the form $U\setminus A$ for some $U\in\tau$, we can show that $\mathcal W\subseteq\tau$ and hence $A=\bigcap\mathcal W$ is a $G_\delta$-set in $Y$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$

Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap_{n\in\omega}W_n$ of some open sets $W_n\in\tau'$. By the choice of the topology $\tau'$, for every $n\in\omega$ there exists open sets $U_n,V_n\in \tau$ such that $W_n=U_n\cup(V_n\setminus A)$. It follows from $A\subseteq W_n=U_n\cup(V_n\setminus A)$ that $A=A\cap W_n=A\cap U_n\subseteq U_n$. $$A=\bigcap_{n\in\omega}W_n=A\cap\bigcap_{n\in\omega}W_n=\bigcap_{n\in\omega}(A\cap W_n)=\bigcap_{n\in\omega}(A\cap U_n)\subseteq \bigcap_{n\in\omega}U_n\subseteq \bigcap_{n\in\omega}W_n=A$$ and hence $A=\bigcap_{n\in\omega}U_n$ is a $G_\delta$-set in $X$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$

Let $\mathfrak q_w$ be the smallest cardinality of a second-countable $T_1$ space which is not a $Q$-space. It is clear that $\mathfrak q_w\le\mathfrak q_0$.

Question. Is $\mathfrak q_w<\mathfrak q_0$ consistent?

Example. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.

Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$

On the other hand we have the following partial affirmative answer to Question 1.

Proof. It is known that Martin's Axiom implies that every second-countable $T_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$

Example 1. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.

Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$

The cardinality $\mathrm{non}(\mathcal M)$ is the above example can be lowered to $\mathfrak q_0$, where $\mathfrak q_0$ is the smallest cardinality of a second-countable metrizable space which is not a $Q$-space (= contains a subset which is not of type $G_\delta$).

A topological space is submetrizable it it admits a continuous metric. Each submetrizable space is functionally Hausdorff in the sense that for any distinct elements $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$.

Example 2. There exists a submetrizable second-countable space $X$ of cardinality $\mathfrak q_0$, which is not symmetrizable.

Proof. By the definition of the cardinal $\mathfrak q_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap\mathcal W$ of a countable family $\mathcal W\subseteq\tau'$. Taking into account that each set $U'\in\tau'\setminus\tau$ is of the form $U\setminus A$ for some $U\in\tau$, we can show that $\mathcal W\subseteq\tau$ and hence $A=\bigcap\mathcal W$ is a $G_\delta$-set in $Y$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$

On the other hand we have the following partial affirmative answer to Question 1.

Proof. It is known that Martin's Axiom implies that every second-countable $T_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$

Let $\mathfrak q_w$ be the smallest cardinality of a second-countable $T_1$ space which is not a $Q$-space. It is clear that $\mathfrak q_w\le\mathfrak q_0$.

Question. Is $\mathfrak q_w<\mathfrak q_0$ consistent?