Let $\mathcal{B} = \{B_1, B_2, \ldots\}$ be a countable basis for $X$.

Part A: $X$ is normal.

Step A1: $X$ is Lindel"of. Let $\{U_\alpha\}$ be an open cover. For each $\alpha$ and each $x \in U_\alpha$, choose $B_{n(x,\alpha)} \in \mathcal{B}$ with $x \in B_{n(x,\alpha)} \subseteq U_\alpha$. The collection $\{B_{n(x,\alpha)}\}$ is a (possibly redundant) subcover indexed by a subset of $\mathbb{N}$, hence countable. For each such $B_k$ in this subcover, choose one $\alpha_k$ with $B_k \subseteq U_{\alpha_k}$. Then $\{U_{\alpha_k}\}$ is a countable subcover.

Step A2: $X$ is normal. This follows from Theorem 7.5: a regular Lindel"of space is normal.

Part B: Constructing separating functions.

Step B1: Define the set of "good pairs": $\mathcal{P} = \{(m, n) \in \mathbb{N}^2 : \overline{B_m} \subseteq B_n\}.$

This is a countable set. Since $X$ is regular: for any $x$ and open $U \ni x$, there exists $B_n \in \mathcal{B}$ with $x \in B_n \subseteq U$, and by regularity there exists an open $V$ with $x \in V \subseteq \overline{V} \subseteq B_n$, and a $B_m$ with $x \in B_m \subseteq V$. So $\overline{B_m} \subseteq \overline{V} \subseteq B_n$, and $(m, n) \in \mathcal{P}$.

Step B2: For each $(m, n) \in \mathcal{P}$, the sets $\overline{B_m}$ and $X \setminus B_n$ are disjoint closed sets (using $\overline{B_m} \subseteq B_n$). By the Urysohn lemma (applicable since $X$ is normal), there exists a continuous function: $g_{m,n}: X \to [0, 1], \quad g_{m,n}|_{\overline{B_m}} = 0, \quad g_{m,n}|_{X \setminus B_n} = 1.$

Enumerate $\mathcal{P}$ as $(m_1, n_1), (m_2, n_2), \ldots$ and write $f_k = g_{m_k, n_k}$.

Part C: The embedding.

Step C1: Define $F: X \to [0, 1]^{\omega}$ by: $F(x) = (f_1(x), f_2(x), f_3(x), \ldots).$

On $[0, 1]^{\omega}$, use the metric: $d(\mathbf{a}, \mathbf{b}) = \sum_{k=1}^{\infty} \frac{|a_k - b_k|}{2^k}.$

This metric induces the product topology on $[0, 1]^{\omega}$.

Step C2: $F$ is injective. Let $x \neq y$. Since $X$ is $T_1$, $\{y\}$ is closed. Set $U = X \setminus \{y\}$. By Step B1, there exists $(m, n) \in \mathcal{P}$ with $x \in B_m \subseteq \overline{B_m} \subseteq B_n \subseteq U$. Then $y \notin B_n$, so $g_{m,n}(x) = 0$ and $g_{m,n}(y) = 1$. Hence $F(x) \neq F(y)$.

Step C3: $F$ is continuous. Each coordinate function $\pi_k \circ F = f_k$ is continuous. By the universal property of the product topology, $F$ is continuous.

Step C4: $F$ is an open map onto $F(X)$. Let $U \subseteq X$ be open and $x \in U$. By Step B1, there exists $(m_k, n_k) \in \mathcal{P}$ with $x \in B_{m_k} \subseteq B_{n_k} \subseteq U$. Then $f_k(x) = 0$ and $f_k(y) = 1$ for all $y \in X \setminus U$.

Consider the set $W = \{\mathbf{a} \in [0,1]^{\omega} : a_k < 1\}$. This is open in $[0,1]^{\omega}$.

We have $F(x) \in W$ (since $f_k(x) = 0$). For any $z \in F^{-1}(W) \cap X$: $f_k(z) < 1$, which means $z \notin X \setminus B_{n_k}$, i.e., $z \in B_{n_k} \subseteq U$.

Thus $F^{-1}(W) \subseteq U$, so $F(U) \supseteq F(X) \cap W$, an open set in $F(X)$ containing $F(x)$.

This shows $F(U)$ is open in $F(X)$.

Conclusion. $F: X \to F(X)$ is a continuous bijection that is also an open map, hence a homeomorphism. Since $F(X) \subseteq [0, 1]^{\omega}$ and $[0,1]^{\omega}$ is metrizable (with metric $d$), $F(X)$ is metrizable, and therefore $X$ is metrizable.