Consider the set $X = [0,1]^{[0,1]}$ with the product topology $\tau_\pi$ and consider it with the supremum metric $d_\infty$ and the induced topology $\tau_d$. Does it hold that $\tau_d = \tau_\pi$ ?
I feel like there is only one inclusion $\tau_d \subset \tau_\pi$ but I hesitate a bit, the product topology feels bigger to me.
First of all by Tychonoff the product is compact, while the sup metric is not. Indeed, for any $r\in[0,1]$ define
$$e_r:[0,1]\to[0,1]$$ $$e_r(x)=\begin{cases} 1 &\text{if }r=x \\ 0 &\text{otherwise} \end{cases}$$
and note that $d_\infty(e_r,e_s)=1$ whenever $r\neq s$. This shows that $d_\infty$ is not compact.
Another difference is that the $[0,1]^{[0,1]}$ under product topology is not first countable, in particular it is not metrizable.
Now about topology inclusions. Consider the open ball $B(0,1/2)$ around the constant $0$ function of radius $1/2$ in $d_\infty$ metric. Clearly it is open. But is it open in the product topology? That would imply that $B(0,1/2)$ contains at least one box of the product topology. And boxes in the product topology have all but finitiely many coordinates covering whole $[0,1]$. But $B(0,1/2)=[0,1/2)^{[0,1]}$. Thus this cannot happen.
What about the inverse inclusion? Let $U:=\prod_{i\in[0,1]}U_i$ be an open box in the product topology, i.e. each $U_i$ is open in $[0,1]$ and $U_i=[0,1]$ for all but finitely many $i$. Let $J\subseteq [0,1]$ be this finite subset such that $U_j\neq [0,1]$ for $j\in J$. Fix $(x_i)\in U$. Then for any $j$ there is $\epsilon_j$ such that $(x_j-\epsilon_j,x_j+\epsilon_j)\subseteq U_j$. In particular if we choose $\epsilon:=\min_{j\in J} \epsilon_j$ (we can do this since $J$ is finite) we get that $B((x_i),\epsilon)$ is fully contained in $U$. By the arbitrary choice of $(x_i)$ we conclude that $U$ is open in the metric topology.
So what we actually get is that $\tau_{\pi}\subsetneq \tau_d$, or in other words that the metric topology is bigger than the product topology.
The supremum metric describes uniform convergence of functions and the product topology pointwise convergence. Therefore $\tau_d \ne \tau_\pi$ . Since uniform convergence implies pointwise convergence, we see $\tau_d$ must be finer than $\tau_\pi$ , i.e. we get $\tau_\pi \subsetneqq \tau_d$ .
The $\tau_d$ metric makes the projections continuous (even contractions) while the product topology is the unique minimal topology that makes the projections continuous so $\tau_\pi \subseteq \tau_d$ is clear and any $\tau_d$ ball of radius $\frac12$ is clearly not $\tau_\pi$ open so the inclusion is strict.
The $\tau_d$ topology is very non-compact and $\tau_\pi$ is compact for another argument.
