Let $\Bbb R^\infty$ be the subset of $\Bbb R^\omega$ (the countably infinite product of $\Bbb R$ with itself) consisting of all sequences that are "eventually zero" that is, all sequences $(x_1, x_2,\dots)$ such that $x_i \ne 0$ for only finitely many values of $i$. Consider the mappings $f_n : \Bbb R^n \to \Bbb R^\omega$ for which $f_n(x)=(x_1,\dots,x_n, 0,0,0,\dots)$. These maps induce a final topology $\mathcal{T}_1$ on $\Bbb R^\infty$. Compare this to the subspace topology $\mathcal{T}_2$ induced on $\Bbb R^\infty$ by $\Bbb R^\omega$.
I know for a fact proven in class that $\mathcal{T}_2 \subseteq \mathcal{T}_1$. And my classmate gave me a hint to consider the box topology in order to find an open set in $\mathcal{T}_1$ that is not in $\mathcal{T}_2$.
Can I just state that since the box topology is finer than the product topology I have that $\mathcal{T}_2 \subsetneq \mathcal{T}_\text{box} \subseteq \mathcal{T}_1$ or is there a possibility that $\mathcal{T}_2 = \mathcal{T}_\text{box}$ in my case? If so then I guess I would need to find an open set in the box topology which isn't in $\mathcal{T}_2$. How can I figure this out?
We have $\mathcal T_2 \subsetneqq \mathcal T_{box}$. As Rob Arthan comments, consider $U = (-1,1)^\omega \cap \mathbb R^\infty$. We have $y \in U$ iff all $y_i \in (-1,1)$. Now assume that $U \in \mathcal T_2$. Since $0 \in U$, there exist $N$ and open $U_i^x \subset \mathbb R$ such that $0 \in V = (\prod_{i=1}^N U^x_i\times \prod_{i=N+1}^\infty \mathbb R) \cap \mathbb R^\infty \subset U$. The point $y = (y_i)$ with $y_i = 0$ for $i \ne N+1$ and $y_{N+1} =1$ is in $V$, but not in $U$.
It seems that you know that $\mathcal T_{box} \subset \mathcal T_1$ which completes the proof.