Is every second-countable Hausdorff space symmetrizable? Let us recall that a symmetric on a set $X$ is any function $d:X\times X\to[0,\infty)$ such that
for every $x,y\in X$ the following two conditions are satisfied:
$\bullet$ $d(x,y)=0$ if and only if $x=y$;
$\bullet$ $d(x,y)=d(y,x)$.
A topological space $X$ is called symmetrizable if there exists a symmetric $d$ on $X$ such that a subset $U\subseteq X$ is open if and only if for every $x\in X$ there exists $\varepsilon>0$ such that $B_d(x,y)\subseteq U$, where $B_d(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}$.
By the Urysohn Metrization Theorem, every regular second-countable space is metrizable and hence symmetrizable.
Question 1. Is each Hausdorff second-countable space symmetrizable?
A weaker version of Question 1 is also of interest:
Question 2. Is each countable first-countable Hausdorff space symmetrizable?

Added in Edit. I have just realized that Question 2 has a simple affirmative answer (which I present below), so only Question 1 remains open.
Added in Edit 31.05.2022. Ups: Question 1 has a negative answer!
 A: I have just realized that the first question has a simple affirmative answer.
Theorem 1. Every countable first-countable $T_1$ space $X$ is symmetrizable.
Proof. For every $x\in X$, fix a neighborhood base $(U_n(x))_{n\in\omega}$ such that $U_{n+1}(x)\subseteq U_n(x)$ for all $n\in\omega$. Since $X$ is countable, there exists a linear order $\le$ on $X$ such that for every $x\in X$ the initial interval $\{y\in X:y\le x\}$ is finite.
It is easy to see that the topology of $X$ is generated by the symmetric
$$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U_n(\min\{x,y\})\big\}.\quad\square$$

Important Remark. After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that affirmative answers to both my questions follow from
Theorem 2.9 (Arhangelski, 1966): Every first-countable $T_1$ space with a $\sigma$-discrete (closed) network is symmetrizable.
But reading the proof of this theorem I discovered that it works only for regular spaces. Arhangelski writes that one loses no generality assuming that the $\sigma$-discrete network is closed, i.e., consists of closed sets. But after taking the closure of elements of a network in a $T_1$-space (even in a Hausdorff space), the network property can be destroyed. So, Question 2 seems to stay open and not answered even for Hausdorff (not mention $T_1$) spaces.

Added in Edit 31.05.2022. To my big surprise (and contrary to what was claimed by Arhangelski in his Theorem 2.9), I have just discovered that Question 1 has negative answer!
First let us prove that the symmetrizability of second-countable spaces is equivalent to the perfectness. Recall that a topological space $X$ is perfect if every closed subset of $X$ is of type $G_\delta$.
Theorem 2. A second-countable (Hausdorff) $T_1$-space $X$ is symmetrizable if (and only if) it is perfect.
Proof. If $X$ is symmetrizable and Hausdorff, then the first-countability of $X$ implies that $X$ is semi-metrizable and perfect, see the paragraph before Theorem 9.8 in Gruenhage's "Generalized metric spaces".
Conversely, if a second-countable $T_1$-space $X$ is perfect, then each open subset of $X$ is an $F_\sigma$ in $X$, which implies that $X$ has a countable closed network. Now we can apply Arhangelski's Theorem 2.9 to conclude that $X$ is symmetrizable. $\square$
Let $\mathrm{non}(\mathcal M)$ denote the smallest cardinality of a nonmeager set in the real line.
Example 1. There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.
Proof. Take any nonmeager linear subspace $L$ of cardinality $\mathrm{non}(\mathcal M)$ in $\mathbb R^\omega$ such that for every $n\in\omega$ the intersection $L_n=L\cap(\{0\}^n\times\mathbb R^{\omega\setminus n})$ is dense in $\{0\}^n\times\mathbb R^{\omega\setminus n}$. Consider the quotient space $X=L_\circ/_\sim$ of $L_\circ=L\setminus\{0\}$ by the equivalence relation $\sim$ defined by $x\sim y$ iff $\mathbb R x=\mathbb Ry$. Since the space $L_\circ$ is Baire and the quotient map $q:L_\circ\to X$ is open, the space $X$ is second-countable and Baire. It is easy to check that the closure of every nonempty set in $X$ contains the set $q[L_n\setminus\{0\}]$ for some $n\in\omega$. This implies that the space $X$ is superconnected in the sense that for every nonempty open sets $U_1,\dots,U_n$ in $X$ the intersection of their closures $\overline U_1\cap\dots\cap\overline U_n$ is not empty.
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 $\bigcap_{n\in\omega}W_n$ 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. $\square$
On the other hand we have the following partial affirmative answer to Question 1.
Theorem 3. Martin's Axiom implies that every second-countable $T_1$ space of cardinality $<\mathfrak c$ is perfect and hence symmetrizable.
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. $\square$
