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
Your proof is fine, except that you want your minimum to run only over $y\in X\setminus\{x\}$.
[Note that this proof (of course) fails for infinite $X$ as the minimum of infinitely many positive numbers need not exist (or, if we switch to the infimum, it need no longer be positive).]