The space is $\mathbb R^\infty$ is defined to be a union (inductive limit) of the chain: $\mathbb R^1$ $\subset$ $\mathbb R^2$ $\subset$ $\mathbb R^3$ $\subset$ ...
The author writes $\mathbb R^\infty$ is the set of sequences $(x_1,x_2,\ldots)$ of real numbers with only finitely many nonzero $x_i$.
The topology in $\mathbb R^\infty$ is defined by the rule: A set F is closed if and only if all the intersections F $\cap$ $\mathbb R^n$ are closed in respective spaces $\mathbb R^n$.
Show that the sequence $(a_1,0,0,\ldots)$, $(0,a_2,0,\ldots)$, $(0,0,a_3,0,\ldots)$, $\ldots$ $(0,\ldots,0,a_n,0,\ldots)$, $\ldots$
has a limit if and only if it has finitely many nonzero terms.
I noticed that if S is the set of terms of the sequence, S is closed in $\mathbb R^\infty$ since its intersection with any $\mathbb R^n$ has finitely many points. So the limit, if it exists, must be of the form L = $(0,\ldots,0,a_k,0,\ldots)$ for some k.
I think I can derive a contradiction if I assume the sequence has infinitely many points, and if I can construct an open set containing L that wouldn't contain infinitely many terms, yet I'm not sure how to do this, since the open sets I can think of all seem to be large and unhelpful.
Is my approach fruitful? I don't have much experience with spaces like $\mathbb R^\infty$, so I assume that F $\cap$ $\mathbb R^n$ is the subset of $\mathbb R^\infty$ obtained from the elements of F by replacing all their entries after the nth entry by 0, and then identified with the corresponding subset of $\mathbb R^n$.
You're right about the interpretation in the final paragraph: the chain of inclusions works with appending $0's$: $\mathbb{R}^n$ as a subset of $\mathbb{R}^\infty$ is $\{(x_n)_n \in \mathbb{R}^{\mathbb{N}^+}: \forall i \ge n+1: x_i = 0\}\subset \mathbb{R}^\infty$, and it gets its topology via the obvious homeomorphism with the usual $\mathbb{R}^n$ (the projection onto the first $n$ coordinates).
We can also define the topology on $\mathbb{R}^\infty$ via open sets, (which follows directly from the definition via closed sets): $O \subseteq \mathbb{R}^\infty$ is open iff $\forall n: O \cap \mathbb{R}^n$ is open in $\mathbb{R}^n$. It's also obvious that all projection maps (mapping a sequence to one coordinate) are continuous (as they are on all $\mathbb{R}^n$), so that convergence in $\mathbb{R}^\infty$ implies convergence in each coordinate space.
So the sequence under consideration is indeed a closed set (the set of its terms), as you rightly note, as its intersection with each $\mathbb{R}^n$ is finite (and hence closed there). But were it to converge, it must converge in each coordinate space and there each sequence is eventually constantly $0$, so the only candidate limit if the all zero-sequence, which thus must lie in the set of terms, so some $a_n$ must be $0$.
Going further, suppose there are infinitely many non-zero $a_i$, so that there is an infinite subset of indices $I$ such that $i \in I$ iff $x_i \neq 0$. Then define the following subset:
$$O = \{(x_n) \in \mathbb{R}^\infty: \forall i \in I: x_i \in (-|a_i|, |a_{i}|), \forall n \notin I : x_n \in (-1,1)\}$$
This is a product of open intervals on each $\mathbb{R}^n$ hence open there, and so $O$ is an open set in $\mathbb{R}^\infty$ which contains $0$, as we saw, the only candidate limit of the sequence. But by construction the set $O$ does not contain any of the terms of the sequence with $a_i$ for a non-zero $a_i$, so it cannot contain all but finitely many terms of the sequence, and so the sequence does not converge to the only candidate $0$, so does not converge at all.
The reverse is trivial: if indeed all but finitely many $a_i$ are $0$, the sequence converges to $0$.