The Hilbert cube $[0,1]^{\mathbb{N}}$ with the product topology is metrizable, by $d(x,y) = \sum_{k} |x_k - y_k| / 2^{\,k+1}$

$\sum_{k} 2^{-(k+1)} = 1$, being $\tfrac{1}{2}\sum_k 2^{-k} = \tfrac{1}{2} \cdot 2$ by [L1] and [L4]; and more generally $\sum_{k \ge N} 2^{-(k+1)} = 2^{-N}$ for every $N \in \mathbb{N}$, by shifting the index.

$2^{-k} \to 0$, since $\sum_k 2^{-k}$ converges by [L1]; so for every real $\eta > 0$ there is $N \in \mathbb{N}$ with $2^{-N} < \eta$.

$d(x,y) = d(y,x)$, term by term, by [L3]. This is (M2).

For $x, y \in Q$ and every $k$: $0 \le |x_k - y_k| \le 1$, since both coordinates lie in $[0,1]$; hence $0 \le |x_k-y_k| 2^{-(k+1)} \le 2^{-(k+1)}$. By [L1] and step 1.1 the series defining $d(x,y)$ converges and $0 \le d(x,y) \le 1$. This is claim 1.

For every $j \in \mathbb{N}$ and $x,y \in Q$: $|x_j - y_j| 2^{-(j+1)} \le d(x,y)$, the left side being a single term of a nonnegative convergent series and hence at most one of its partial sums, which is at most the sum by [L1].

$d(x,z) \le d(x,y) + d(y,z)$: for every $k$ one has $|x_k-z_k| \le |x_k-y_k| + |y_k-z_k|$ by [L3], so each partial sum of the left series is at most the corresponding partial sum of the sum of the two right series by [L2], and the inequality passes to the limits by [L2] and [L1]. This is (M3), so claim 2 holds.

Every $d$-ball contains a basic product-open neighbourhood of its centre. Given $u \in Q$ and $\rho > 0$, take $N$ with $2^{-N} < \rho/2$ by step 1.2 and put $W := \{\, y \in Q : |y_k - u_k| < \rho/2 \text{ for every } k < N \,\}$, which is a basic product-open set containing $u$ by [A1] and [A2]. For $y \in W$, splitting the series at $N$ gives $d(u,y) \le (\rho/2)\sum_{k<N}2^{-(k+1)} + \sum_{k \ge N} 2^{-(k+1)} \le \rho/2 + 2^{-N} < \rho$, by steps 1.1 and 2.1 with [L1] and [L2]. So $u \in W \subseteq B_d(u,\rho)$.

$d(x,x) = 0$, every term vanishing by [L3]. Conversely if $d(x,y) = 0$ then by step 3.1 every $|x_j-y_j|2^{-(j+1)} \le 0$, so $|x_j - y_j| = 0$ and $x_j = y_j$ for every $j$ by [L3] and [L4]; hence $x = y$. This is (M1).

Every subbasic product-open set is $d$-open. Let $O$ be open in $[0,1]$, let $x \in \pi_j^{-1}[O]$ and take $\varepsilon > 0$ with $\{\, t \in [0,1] : |t - x_j| < \varepsilon \,\} \subseteq O$, available by [A2]. If $d(x,y) < \varepsilon 2^{-(j+1)}$ then $|x_j-y_j| 2^{-(j+1)} < \varepsilon 2^{-(j+1)}$ by step 3.1, so $|x_j - y_j| < \varepsilon$ by [L4], so $y_j \in O$; hence $B_d(x, \varepsilon 2^{-(j+1)}) \subseteq \pi_j^{-1}[O]$.

$\mathcal{T}_d$ is contained in the product topology: by [L6] it suffices that every ball $B_d(x,r)$ be product-open, and for $z \in B_d(x,r)$ the triangle inequality of step 3.2 gives $B_d(z, r - d(x,z)) \subseteq B_d(x,r)$ with $r - d(x,z) > 0$, while step 3.3 supplies a basic product-open $W$ with $z \in W \subseteq B_d(z, r - d(x,z))$.

The product topology is contained in $\mathcal{T}_d$: by step 4.2 every subbasic product-open set is $d$-open, and $\mathcal{T}_d$ is a topology containing them, hence contains the topology they generate, which is the product topology by [A1].

By steps 5.1 and 4.3 the metric topology of $d$ is the product topology on $Q$, so $Q$ is metrizable; this is claim 3, and with steps 2.1 and 3.2 all three claims are proved.