I want to show the opposite containment of basis, i.e. $B_\text{box} \subseteq B_\text{unif} \subseteq B_\text{prod} $.
Starting with $B_\text{unif} \subseteq B_\text{prod} $, the basic open set of $B_\text{unif} = B_\epsilon(x)$ where $\epsilon$ is generated by $d(x,y) = \sup\{ d_1= \min\{1,| x-y|\}\}$ and basic open set of $B_\text{prod} = \left\{ \prod_{n \in \Bbb{N}} U_n : U_n = \Bbb{R}\ \text{ for all but finitely many } n\right\}$, then pick a $y_n$ in $B_\epsilon(x)$, we want to show that $y_n$ is also in $\prod U_n$ we know that there exists a finite set $F$, s.t $U_n = \Bbb{R} \ \forall\ n \in \Bbb{N} \setminus F$ for $y_n$ s.t $n$ is in $\Bbb{N} \setminus F$ then $y_n$ is obviously $n$ $U_n$, for $n$ in $F$, we know that by metrizability of $\Bbb{R}$ there exit a $C_{\epsilon_n}(y) \subseteq U_n$, so pick $\epsilon = \frac{1}{2} \min \{ \epsilon_n : n \in F \}$ which we can do since $F$ is finite then for $n$ in $F$: $d_1(x_n, y_n) \leq \sup(d_1(x_k, y_k))$ for $k$ in $\Bbb{N} = d \leq \epsilon \leq \epsilon_n$ and hence $B_\epsilon(x) \subseteq \prod U_n$ as required. For containment between $B_\text{box} \subseteq B_\text{unif}$ $B_\text{box} = \prod U_n$: s.t $U_n $is open in $\Bbb{R}$. so pick each $U_n = B_{\epsilon_n}$ where $\epsilon_n$ is generated by $d_1$ which we can do by metrizability of $\Bbb{R}$, then $\prod U_n \subseteq B_\epsilon(x)$ since $d_1 \leq d = \sup(d_1)$
I also want to show strict containment so for instance $ (0,1) \times (3,5) \times (1,2)\times\cdots$ is a set of $T_\text{box}$ but not $T _\text{unif}$ and $(2,0) \times (2,0)\times\cdots$ is an open set of $T_\text{unif}$ but not $T_\text{prod}$.
I'm not sure if I can pick $\epsilon$ since shouldn't it just be supremum of all the $\epsilon_n$ generated by $d_1$ I'm sorry my thoughts with this proof are all over the place, if someone can help me understand it better I would really appreciate it.