For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space

For $y \in \mathbb{R}^n$: $d_\infty(x,y) < r$ if and only if $|x_k - y_k| < r$ for every $k < n$, since by [L3] the maximum is one of the values $|x_k-y_k|$ and is an upper bound for all of them.

For $t \in \mathbb{R}$ and $r > 0$: $|t - y_k| < r$ if and only if $y_k \in (t-r, t+r)$, by [L1].

Write $t_k := |x_k - y_k|$ and $M := d_\infty(x,y) = \max\{t_k : k<n\}$. Then $t_j \le M$ for every $j < n$ and $M = t_{j_0}$ for some $j_0 < n$, by [L3].

$d_2(x,y)^2 = \sum_{k<n}(x_k-y_k)^2$ by [L5], and $(x_k-y_k)^2 = t_k^2 \ge 0$ by [L5].

$n \le n^2$ as reals: for $n \ge 1$ the canonical natural satisfies $\iota(1) = 1 \le \iota(n)$ by [L7], so either $\iota(n) = 1$, in which case $\iota(n) = \iota(n)^2 = 1$, or $1 < \iota(n)$, in which case multiplying that strict inequality by $\iota(n) > 0$ gives $\iota(n) < \iota(n)^2$ by [L7].

Conversely let $B = \prod_{k<n} U_k$ be a box with every $U_k$ open in $\mathbb{R}$ and let $x \in B$. For each $k<n$ the set $\{\, \rho \in \mathbb{R} : \rho > 0,\ (x_k-\rho,\ x_k+\rho) \subseteq U_k \,\}$ is nonempty by [L1], so [L6] supplies $\rho_k$ in it for every $k<n$; put $r := \min\{\rho_k : k<n\}$, which exists and is positive by [L3].

$d_1(x,y) = \sum_{k<n} t_k \le \sum_{k<n} M = n M$, using $t_k \le M$ from step 1.3 and [L4].

$M = t_{j_0} \le \sum_{k<n} t_k = d_1(x,y)$, since every $t_k \ge 0$ by [L5] and a single nonnegative term is at most the sum, by [L4].

$B_{d_\infty}(x,r) = \prod_{k<n}(x_k - r,\ x_k + r)$: by step 1.1 a point $y$ lies in the ball exactly when $|x_k-y_k| < r$ for every $k<n$, and by step 1.2 that says exactly $y_k \in (x_k-r, x_k+r)$ for every $k < n$.

$M^2 = t_{j_0}^2 \le \sum_{k<n} t_k^2 = d_2(x,y)^2$ by steps 1.3 and 1.4 with [L4], and both $M$ and $d_2(x,y)$ are nonnegative by [L2] and [L5], so $M \le d_2(x,y)$ by [L5].

$d_2(x,y)^2 = \sum_{k<n} t_k^2 \le \sum_{k<n} M^2 = n M^2 \le n^2 M^2 = (nM)^2$, using $t_k \le M$ with [L5] and [L4], then step 1.5 with $M^2 \ge 0$; since $d_2(x,y) \ge 0$ and $nM \ge 0$, [L5] gives $d_2(x,y) \le n M$.

Every $d_\infty$-ball is a box with open factors, by step 2.3 and [L1], hence a basic open set of the product topology by [A2]; so every $d_\infty$-open set is product-open, by [L2] and [A2].

With $r$ as in step 1.6: $B_{d_\infty}(x,r) = \prod_{k<n}(x_k-r, x_k+r) \subseteq \prod_{k<n}(x_k-\rho_k, x_k+\rho_k) \subseteq B$, since $r \le \rho_k$ for every $k$ by [L3].

Steps 2.1, 2.2, 2.4 and 2.5 give $d_\infty \le d_1 \le n\,d_\infty$ and $d_\infty \le d_2 \le n\,d_\infty$ at every pair of points, which is claim 2, the constants $1$ and $n$ being positive by [L7].

By steps 1.6 and 3.2 every basic open set of the product topology is $d_\infty$-open by [L2], hence every product-open set is $d_\infty$-open; with step 3.1 this gives claim 1.

By step 3.3 and [L8] the metrics $d_1$, $d_2$ and $d_\infty$ have the same metric topology, which by step 4.1 is the product topology; so all three induce it and $\mathbb{R}^n$ with the product topology is metrizable. This is claim 3, and with steps 4.1 and 3.3 all three claims are proved.