Here is Prob. 3, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:
Let $X_n$ be a metric space with metric $d_n$, for $n \in \mathbb{Z}_+$.
(a) Show that $$ \rho ( \mathbf{x}, \mathbf{y} ) = \max \left\{ \ d_1 \left( x_1, y_1 \right), \ldots, d_n \left( x_n, y_n \right) \ \right\} $$ is a metric for the product space $X_1 \times \cdots \times X_n$.
(b) Let $\bar{d}_i = \min \left\{ \ d_i, 1 \ \right\}$. Show that $$ D ( \mathbf{x}, \mathbf{y} ) = \sup \left\{ \ \bar{d}_i \left( x_i, y_i \right) / i \ \right\} $$ is a metric for the product space $\prod X_i$.
Here is the link to my Math SE post on Part (a).
So here I will only be attempting Part (b).
My Attempt:
Part (b)
Let $B \colon= \prod_{i=1}^\infty U_i $ be an arbitrary basis element for the product topology on $\prod_{i=1}^\infty X_i$, where, for each $i \in \{ \ 1, 2, 3, \ldots, \ \}$, the set $U_i$ is a subset of $X_i$ such that $U_i$ is open in $\left( X_i, d_i \right)$, and $U_i \not= X_i$ for at most finitely many values of $i$; let $i \in \left\{ \ i_1, \ldots, i_n \ \right\}$ be the values of $i$ for which $U_i \neq X_i$. Let $\mathbf{p} \colon= \left( p_1, p_2, p_3, \ldots \right)$ be an arbitrary element of $B$.
Then, for each $i \in \left\{ \ i_1, \ldots i_n \ \right\}$, $p_i \in U_i$ and $U_i$ is open in the metric space $\left( X_i, d_i \right)$, which implies that there is a real number $\delta_i > 0$ such that the set $$B_{d_i} \left( p_i, \delta_i \right) \colon= \left\{ \ x \in X_i \colon \ d_i \left( x, p_i \right) < \delta_i \ \right\}$$ is contained in $U_i$. Let us assume that $\delta_i \in (0, 1)$ for each $i \in \left\{ \ i_1, \ldots, i_n \ \right\}$.
Let $$\delta \colon= \min \left\{ \ \frac{\delta_{i_1}}{i_1}, \ldots, \frac{\delta_{i_n}}{i_n} \ \right\}.$$ Then $\delta > 0$ of course.
Moreover, if $\mathbf{x} \colon= \left( x_1, x_2, x_3, \ldots \right)$ is any point in $\prod_{i=1}^\infty X_i$ such that $$\mathbf{x} \in B_{D} ( \mathbf{p}, \delta ) \colon= \left\{ \ \mathbf{y} \in \prod_{i=1}^\infty X_i \ \colon \ D ( \mathbf{y}, \mathbf{p} ) < \delta \ \right\}, $$ then for each $i \in \{ \ i_1, \ldots, i_n \ \}$, we have $$ \frac{\bar{d}_i \left( x_i, p_i \right)}{i} \leq D ( \mathbf{x}, \mathbf{p} ) < \delta \leq \frac{\delta_i}{i}, $$ and so $$ \bar{d}_i \left( x_i, p_i \right) < \delta_i < 1, $$ and hence $$ d_i \left( x_i, p_i \right) = \bar{d}_i \left( x_i, p_i \right) < \delta_i,$$ which implies that $x_i \in B_{d_i} \left( p_i, \delta_i \right)$ and hence $x_i \in U_i$. So $\mathbf{x} \in B$. Moreover, $\mathbf{p} \in B_{D} ( \mathbf{p}, \delta)$.
Thus, we have shown that, for any basis set $B$ for the product topology on $\prod_{i=1}^\infty X_i$ and for any element $\mathbf{p} \in B$, there is a basis element $B^\prime \colon= B_{D} ( \mathbf{p}, \delta)$ for the $D$-metric topology on $\prod_{i=1}^\infty X_i$ such that $$ \mathbf{p} \in B^\prime \subset B. $$ Therefore the $D$-metric topology is finer than the product topology on $\prod_{i=1}^\infty X_i$.
Is this part of the proof correct? Is the presentation easy to follow?
Now let $B_{D}(\mathbf{p}, \epsilon )$ be an arbitrary basis element for the $D$-metric topology on $\prod_{i=1}^\infty X_i$, where $\mathbf{p} \colon= \left( p_1, p_2, p_3, \ldots \right)$ is an arbitrary point of $\prod_{i=1}^\infty X_i$ and $\epsilon$ is a positive real number.
Now for each point $\mathbf{x} \in B_{D} ( \mathbf{p}, \epsilon)$, we need to find a basis element $B_{\mathbf{x}}$ for the product topology on $\prod_{i=1}^\infty X_i$ such that $$\mathbb{x} \in B_{\mathbf{x}} \subset B_{D}( \mathbf{p}, \epsilon). \tag{1} $$
Let $\mathbf{x} \colon= \left( x_1, x_2, x_3, \ldots \right)$ be an arbitrary point in $B_D ( \mathbf{p}, \epsilon)$. Then $\mathbf{x} \in \prod_{i-1}^\infty X_i$ and $D( \mathbf{x}, \mathbf{p} ) < \epsilon$; so for any real number $\delta$ such that $0 < \delta < \epsilon - D( \mathbf{x}, \mathbf{p})$, we have the open ball $B_D ( \mathbf{x}, \delta)$ such that $$ B_D ( \mathbf{x}, \delta) \subset B_D ( \mathbf{p}, \epsilon ). \tag{2} $$
Let $n$ be a natural number such that $n > 2/\delta$, and then let $$ B_{\mathbf{x}} \colon= B_{d_1} \left( x_1, \frac{\delta}{2} \right) \times \cdots \times B_{d_n} \left( x_n, \frac{\delta}{2} \right) \times X_{n+1} \times X_{n+2} \times \cdots. \tag{A} $$ As every open ball in a metric space is an open set, so this set $B_{\mathbf{x}}$ is a basis element for the product topology on $\prod_{i=1}^\infty X_i$; and $\mathbf{x} \in B_{\mathbf{x}}$ of course.
Moreover, if $\mathbf{y} \colon= \left( y_1, y_2, y_3, \ldots \right) \in B_{\mathbf{x}}$, then for each $i \in \{ \ 1, \ldots, n \ \}$, we have $$ \frac{ \bar{d}_i \left( y_i, x_i \right) }{ i } \leq \bar{d}_i \left( y_i, x_i \right) \leq d_i \left( y_i, x_i \right) < \frac{\delta}{2}; \tag{3} $$ and for $i \in \{ \ n+1, n+2, n+3, \ldots \ \}$, we have $$ \frac{ \bar{d}_i \left( y_i, x_i \right) }{ i } \leq \frac{1}{i} \leq \frac{1}{n} < \frac{\delta}{2}. \tag{4} $$ Since we have chosen $n$ to be a natural number such that $n > 2/\delta$, so we have the inequality $$ \frac{1}{n} < \frac{\delta}{2},$$ which we have used in (4).
From (3) and (4) we can conclude that $$ D ( \mathbf{y}, \mathbf{x} ) = \sup \left\{ \ \frac{ \bar{d}_i \left( y_i, x_i \right) }{i} \ \colon \ i \in \mathbf{N} \ \right\} \leq \frac{\delta}{2} < \delta, $$ which shows that $\mathbf{y} \in B_D ( \mathbf{x}, \delta)$ and hence $ \mathbf{y} \in B_D( \mathbf{p}, \epsilon)$, by virtue of (2) above.
Since $\mathbf{y}$ was an arbitrary point of the set $B_{\mathbf{x}}$ as defined in (A) above, therefore we can conclude $B_{\mathbf{x}} \subset B_D ( \mathbf{p}, \epsilon)$, and that (1) holds with this set $B_{\mathbf{x}}$.
So the product topology on $\prod_{i-1}^\infty X_i$ is finer than the $D$-metric topology.
Is this part of the proof the right one in terms of its logic and presentation?
Hence these two topologies on $\prod_{i=1}^\infty X_i$ are the same.
Is my proof sound enough? Is every step in both parts of this proof correct? Is the presentation accessible enough too, especially for a student who might be studying topology for the first time?