Let $\Bbb R^\infty$ be the set of sequences in $\Bbb R^{\Bbb N}$ for which $x_j \ne 0$ for finitely many $j$. The maps $f_n:\Bbb R^n \to \Bbb R^{\Bbb N}$ where $f_n(x)=(x_1,x_2, \dots x_n,0,0 \dots)$ and $n \in \Bbb N$ induce the final topology $\tau$ on $\Bbb R^{\infty}$. Compare this topology with the subspace topology inherited from $\Bbb R^{\Bbb N}$.
By definition the final topology $\tau$ is the finest topology you can put on $\Bbb R^\infty$ such that $f_n$’s are continuous. Therefore if $\tau’$ is the subspace topology inherited from $\Bbb R^{\Bbb N} $ we have that $$\tau’ \subseteq \tau.$$
I’m stuck with the other inclusion. I think that it’s by definition that $$\tau’=\{\Bbb R^\infty \cap O \mid O \text{ open in } \Bbb R^{\Bbb N} \}$$ and that $$\tau=\{V \subset \Bbb R^{\Bbb N} \mid f^{-1}_n(V) \text{ open in } \Bbb R^{n} \}.$$
Now if I’m to show that $\tau \subseteq \tau’$, then picking $O \in \tau$ I have that $$f^{-1}_n(O)=\{x \in \Bbb R^n \mid (x_1,x_2,\dots, x_n,0,0, \dots) \in \Bbb R^{\Bbb N} \}$$ is open in $\Bbb R^n$. How can I see wheter this set belongs to $\tau’$?
If $U \in \tau’$, then $$U=\Bbb R^\infty \cap V$$ where $V$ is open in $\Bbb R^{\Bbb N} $. This can be expressed as $$U= \Bbb R^\infty \cap V = \Bbb R^\infty \cap (\Bbb R \times \Bbb R \times \dots\times V_{i_1} \times V_{i_2} \times \dots \times V_{i_m} )$$ for some $m \in \Bbb N$. What can I do know? Am I to look at the preimage $$f^{-1}_n(\Bbb R^\infty \cap (\Bbb R \times \Bbb R \times \dots\times V_{i_1} \times V_{i_2} \times \dots \times V_{i_m} ) )= f^{-1}_n(\Bbb R^\infty) \cap f^{-1}_n(\Bbb R \times \Bbb R \times \dots\times V_{i_1} \times V_{i_2} \times \dots \times V_{i_m} )?$$
It looks as $$f^{-1}_n(\Bbb R \times \Bbb R \times \dots\times V_{i_1} \times V_{i_2} \times \dots \times V_{i_m} ) =\{x \in \Bbb R^n \mid (x_1,x_2,\dots, x_n,0,0, \dots) \in \Bbb R \times \Bbb R \times \dots\times V_{i_1} \times V_{i_2} \times \dots \times V_{i_m} \}$$
but I don’t think this helps. Any ideas what to do here?
My comments where a little bit misleading but indeed we can show that $\tau' \subsetneq \tau$.
Because each $f_n$ is continuous if we consider in $\mathbb{R}^\mathbb{N}$ the product topology, then each $f_n$ is continuous when we consider its range in $\mathbb{R}^\infty$ with the restricted topology $\tau'$.
Then, as you said, the final topology $\tau$ is the finest topology you can put on $\mathbb{R}^\infty$ such that each $f_n$ is continuous, and so we get that $\tau' \subset \tau$.
Now, to prove that this topologies are not equal consider the set $$V=\mathbb{R}^\infty \cap \prod_{n \in \mathbb{N}} (-1,1)$$ Observe that clearly $V \subset \mathbb{R}^\infty$ and $$f_n^{-1}(V)=\{(y_1, \cdots, y_n) \in \mathbb{R}^n: -1 <y_i<1, i=1, \cdots, n\} = (-1,1)^n$$ for each $n \in \mathbb{N}$, so we have that $V \in \tau$.
Now, let's see that $V \not \in \tau'$.
Suppose that $V \in \tau'$, then as $x=(1/2,0,0,\cdots) \in V$ there exists a basic open set $U \in \tau'$ such that $x \in U \subset V$.
As $U$ is a basic set for the product topology on $\mathbb{R}^\infty$, we have that $$U=\mathbb{R}^\infty \cap \prod_{n \in \mathbb{N}} U_n$$ where every $U_n$ is open in $\mathbb{R}$, and there exists an $n_0 \in \mathbb{N} $ such that $U_n = \mathbb{R}$ for every $n \geq n_0$.
Now, as $x \in U$, we have that $1/2 \in U_1$ and $0 \in U_n$ for every $n >1$.
So define $y: \mathbb{N} \longrightarrow \mathbb{R}$ given by
$$y_n= \left\{ \begin{array}{lcc} 1/2 & \text{if} & n=1 \\ \\ 0 & \text{if} & 1<n<n_0 \\ \\ 2 & \text{if} & n =n_0\\ \\ 0 & \text{if} & n>n_0 \end{array} \right.$$ and we get that $y \in U$ but $y \not \in V$, so we have the desired contradiction.