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.
my guess here just consider the topology induced by the metric as well
as the product topology and show that they are included in each other.
Yes, exactly. As you noted, if $T$ is the product topology then an open set in $T$ is of the form $U \times V$ where $U$ is open in $(M,d)$ adn $V$ is open in $(N,d')$. To show $T=T_1$ you can show that
(i) for $(x,y) \in U \times V$ there exists $ \delta$ such that $B_{d_1}((x,y),\delta ) \subseteq U \times V$
and
(ii) If $B_{d_1}((x,y), r)$ is an open ball (in $T_1$) and $(x',y') \in B_{d_1}((x,y), r)$ then there exists $U \times V$ in the product topology such that $(x',y') \in U \times V \subseteq B_{d_1}((x,y), r)$
The same two proofs with $d_1$ replaced by $d_2$ will yield $T = T_2$.
Now regarding $T = T_1$:
$(1a)$ One of the ideas to solve this problem was to set
$r<\displaystyle\frac{1}{2}\operatorname{min}\{\delta,\epsilon\}$.
Doing this I believe that from $d(x,y)+d'(x',y')<r$ I can conclude
that $d(x,y)<\delta$ and $d'(x',y')<\epsilon$, therefore the ball is
included in the set.
Yes that's also correct. This appears to be direction (ii) above:
Let $U \times V$ be open in $T$ and $(x,y) \in U \times V$ and consider $B((x,y),r)$ where $r < {\min (\delta , \varepsilon) \over 2}$. Then for any $(x',y') \in B((x,y),r)$ we have
$$ d_1 ((x,y), (x',y')) = d(x,x') + d(y,y') < r$$
In particular, $d(x,x') < \delta$ and $d(y,y') < \varepsilon$ hence $x' \in U$ and $y' \in V$ and hence $(x',y') \in U \times V$, showing that $B((x,y),r) \subseteq U \times V$ hence $U \times V$ is open in $T_1$.
(2a)...
From the above you can see that $r = \delta + \varepsilon$ does not work.
Now I will assume the open ball is given, how can I pick $\delta$ and
$\epsilon$ to have $U\times V$ inside the ball? Taking
$\delta=\epsilon < \frac{1}{2}r$ seems to be enough.
Does this prove that $T=T_1$?
No but almost. Let $B_{d_1}((x,y),r)$ be an open ball in $T_1$ and $(x',y') \in B((x,y),r)$. You want to find an open set in $T$ that contains $(x',y')$ and is contained in $B((x,y),r)$. Let $R = {\min (r-d_1((x,y),(x',y'), d_1((x,y),(x',y'))\over 2}$ and consider the set $B_d(x', R) \times B_{d'}(y',R)$ (open in $T$). Then $d_1((x,y),(x',y')) = d(x,x') + d(y,y') < \min( r-d_1((x,y),(x',y'), d_1((x,y),(x',y'))) < r$.
(2)...
(2) is easier than (1): Note that if $\max (\delta, \varepsilon) < r$ then both $\delta < r$ and $\varepsilon < r$.
I hope this helps and if not leave me a comment.
Best Answer
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:
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.