I'll use the $\rho_n$ that are bounded by 1, and $\rho(x,y) = \sum_n \frac{\rho_n(x_n,y_n)}{2^n}$ metric on the product $X = \prod_n X_n$.
Let $O$ be a basic open product set, so $O = \prod_n O_n$, all $O_n$ are open in $X_n$ and where we have a finite set $F \subset \mathbb{N}$ such that $n \notin F$ iff $X_n = O_n$. We want to show it is open in the $\rho$-topology, so pick $x \in O$, and we want to find $r>0$ such that $B_{\rho}(x, r) \subset O$. This would show that all basic product open sets are $\rho$-open, and thus all product open sets are $\rho$-open.
Now, for every $n \in F$, we have that $x_n \in O_n$, which is a (non-trivial) open subset in $X_n$, so we have $r_n > 0$ such that $B_{\rho_n}(x_n, r_n) \subset O_n$, from the fact that the topology on $X_n$ is induced by the metric $\rho_n$. As we have finitely many $r_n$ to consider, we can find $0 < r < 1$ such that $r \le \frac{r_n}{2^n}$ for all $n \in F$.
The claim now is that this $r$ is as required, in the sense that $B_{\rho}(x, r) \subset O$.
To see this, take any $y$ with $\rho(x,y) < r$. For $n \in F$, we know that $\frac{\rho_n(x_n, y_n)}{2^n} \le \rho(x, y) < r \le \frac{r_n}{2^n}$, which implies that for such $n$ we have that $\rho_n(x_n, y_n) < r_n$, and so $y_n \in B_{\rho_n}(x_n, r_n) \subset O_n$. Hence, for all $n \in F$, $y_n \in O_n$, and as the other $O_n$ equal $X_n$ by the form of $O$, we have that indeed $y \in O$, and as $y$ was arbitrary, $B_\rho(x, r) \subset O$, as required.
Now for the other part: we start with an open ball $B_\rho(x,r)$, a basic open subset of the $\rho$-topology, for some arbitrary $x \in X$ and $r>0$, and try to find a basic open subset in the product topology $O$ such that $x \in O \subset B_\rho(x,r)$. This would then show that any $\rho$-open ball is open in the product topology and would show the other inclusion we need: every $\rho$-open set is product open.
The intuition is that the tail of a series like the one that defines $\rho$ is essentially irrelevant (we can get it as small as we like) and this corresponds to the idea that basic open subsets only depend on finitely many non-trivial open sets. So we first pick $N \in \mathbb{N}$ such that $\frac{1}{2^N} < \frac{r}{2}$. This $N$ defines our tail. For $1 \le k \le N$ we consider the open balls $O_k = B_{\rho_k}(x_k, \frac{r}{2N})$, and we set $O_k = X_k$ for $k \ge N+1$.
The claim now is that $O = \prod_k O_k \subset B_\rho(x, r)$, as required. Note that $O$ is indeed a basic open subset in the product topology on $X$ and $x \in O$. To verify the latter claim, we simply estimate: let $y$ be in $O$, then for $k \le N$, $\rho_k(x_k, y_k) < \frac{r}{2N}$, so $$\sum_{k=1}^{N} \frac{\rho_k(x_k,y_k)}{2^k} \le \sum_{k=1}^{N} \rho_k(x_k,y_k) < N\cdot \frac{r}{2N} = \frac{r}{2}\mbox{,}$$ while $$\sum_{k=N+1}^{\infty} \frac{\rho_k(x_k, y_k)}{2^k} \le \sum_{k=N+1}^{\infty} \frac{1}{2^k} = \frac{1}{2^N} < \frac{r}{2}\mbox{.}$$
Putting it together, we indeed get that for $y \in O$ we have $\rho(x,y) < \frac{r}{2} + \frac{r}{2} = r$, as required.
Note that if $(X,d)$ is metric space, then $d'=d/(1+d)$ generates same topology of $(X,d)$. So we only prove this proposition:
Let $(X_n,d_n)$ be a sequence of metric spaces, and $d_n(x,y)\le 1$ for all $n$ and $x,y\in X_n$, then $d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n)$ generates the product topology of $X=\prod_n X_n$.
At first, we prove that for each $a=(a_n)_{n=1}^\infty\in X$ and $r>0$, there is a open basis $V$ of $X$ satisfy that $a\in V\subset B_d(a,r)$.
Let $N$ be a natural number that satisfy $2^{-N}\le r/2$. Consider
$$
V= B_1 (a_1,r/2)\times \cdots\times B_N (a_N,r/2)\times X_{N+1}\times X_{N+2}\times\cdots
$$
(where $B_i(x,r)$ is a open ball in $X_i$.) If $x\in V$ then $d(a_i,x_i)<r/2$ for
$i=1,2,\cdots, N$. Therefore
$$
\begin{aligned}
d(a,x)=\sum_{n=1}^\infty 2^{-n}d_n(a_n,x_n)&\le \frac{1}{2}\sum_{n=1}^N 2^{-n}r + \sum_{n>N}2^{-n}\\
&=(1-2^{-N})\cdot\frac{r}{2}+2^{-N}\\
&< \frac{r}{2}+\frac{r}{2}=r
\end{aligned}
$$
so $x\in B_d(a,r)$.
Finally, you prove that for each $a\in X$ and for each basis $V$ of $X$ that contaning $a$
there is $r$ satisfy that $a\in B_d(a,r)\subset V$. It is easy to prove so I leave the proof of this part for yours.
Best Answer
From the geometry, it is sort of obvious that $B_{d}\left(x, \epsilon\right) \subset B_{p}\left(x, \epsilon\right)$
However, you asked for a proof. First let us show $y \in B_{d}\left(x, \epsilon\right) \Rightarrow y \in B_{p}\left(x, \epsilon\right)$:
$$ y \in B_{d}\left(x, \epsilon\right) \\ \Leftrightarrow d(x, y) < \epsilon \\ \Rightarrow p(x, y) < \epsilon \\ \Leftrightarrow y \in B_{p}\left(x, \epsilon\right) $$
where the third line is the inequality $p(x, \epsilon) \le d(x, \epsilon)$. To show that $y \in B_{p}\left(x, \epsilon\right) \nRightarrow y \in B_{p}\left(x, \epsilon\right)$, consider the point $y=\left( x_1 + \alpha\epsilon, x_2 + \alpha \epsilon, \dots \right)^T$, then $p(x, y)=\alpha\epsilon$, while $d(x, y)=\sqrt{n}\alpha\epsilon$, so for $\frac{1}{\sqrt{n}} < \alpha < 1$, $y \not\in B_d\left(x, \epsilon\right)$.