Prob. 3 (b), Sec. 21, in Munkres' TOPOLOGY, 2nd ed: A Countably Infinite Cartesian Product Of Metrizable Spaces Is Metrizable 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). 
Prob. 3 (a), Sec. 21, in Munkres' TOPOLOGY, 2nd ed: A Finite Cartesian Product Of Metrizable Spaces Is Metrizable
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? 
 A: Actually I am a student who's studying topology for the first time, and if it is still relevant I would like to express my opinion, so you can get an aswer from a student's point of view.
Your proof seems to be correct, and it is worth noting that detailed proofs and explanations are always helpful for beginners, at least for me - So thanks!
I really appreciate your intention to be as clear as possible for your students.
On the other hand, if a proof becomes too lengthy it is sometimes hard to follow.
Also I would like to point out that in my opinion, the more someone encounters different methods to tackle a problem, the more likely the next time he/she encounters a similar problem their set of tools will be larger, and hence their chances to solve the problem will increase.
In both directions of your proof you used the same approach. You took an arbitrary element and a basic set which contains it in one topology and then showed that there exists a basic set in the other topology that contains the element and is contained in the basic set of the first topology.
I would like to suggest a different approach for the second direction, which I believe is also shorter. I am going to tweak a tiny bit Munkres' proof for the specific case about $\mathbb R ^ \omega$ (theorem 20.5), and as you will notice it uses pretty much the same elements your proof does, but as I mentioned the approach is different:
Alternative proof for second direction:
Let $U$ be an arbitrary open set in the D-metric topology, and let $x$ be an arbitrary element in $U$. Since $U$ is open in a metric space we have that there exists $\varepsilon >0$ such that $B_D(x,\varepsilon) \subseteq U$, and there exists $N \in \mathbb N$ such that $\frac {1}{N} < \varepsilon$.
Let us define $V_x=B_{d_1}(x,\varepsilon) \times B_{d_2}(x,\varepsilon)\times \dots \times B_{d_N}(x,\varepsilon) \times X_{N+1} \times X_{n+2} \times \dots$ and notice that $V_x$ is a basic set in the product topology and hence open.
If $y\in \prod \limits _{n=1} ^{\infty} X_n$ then for each $n > N$ $y_n$ satisfies:
$ \frac {1}{n} \overline d _n (x_n,y_n) \overset {\overline d _n (x_n,y_n) =min\{1,d_n\}} {\le} \frac {1}{n} \cdot 1 < \frac {1}{N} $
Also, let us denote $M_y = \underset {n\in [N]}{max}\left\{ \frac {1}{n} \overline d _n (x_n,y_n) \right\}$.
So we can conclude that for each $n\in \mathbb N$ we have $\frac {1}{n} \overline d _n (x_n,y_n) \le max \left\{\frac {1}{N}, M_y \right\}$ which implies that $D(x,y) \le  max \left\{\frac {1}{N}, M_y \right\}$.
We will show now that $V_x \subseteq B_D(x,\varepsilon) \subseteq U$:
If $y\in V$ we have that for each $n \le N$: $\quad d_n(x_n,y_n) < \varepsilon \overset {\overline d _n (x_n,y_n) =min\{1,d_n\}}{\implies} \frac {1}{n} \overline d _n (x_n,y_n) < \varepsilon \implies M_y <\varepsilon$
And every $y$ satisfies $ \frac {1}{n} \overline d _n (x_n,y_n) < \frac {1}{N} $ for $n>N$, and remember that $\frac {1}{N} < \varepsilon$.
Combining both inequalities we get $D(x,y) \le max\left\{\frac {1}{N}, M_y \right\} < \varepsilon \implies y \in B_D(x,\varepsilon) \subseteq U$.
Finally, notice that $U = \bigcup \limits _{x\in U} V_x$, and as $U$ is a union of open sets in the product topology we have that $U$ is open in the product topology.
End of proof
To sum it all up, I think this technique of showing that a set is open by finding a way to present it as a union of open sets (or in other words finding an open neighborhood for each point which is contained in the set) is very useful, and if you show this question to your students it is worth mentioning this method.
I hope this was helpful for you.
