A compact subset of $\mathbb{R}^\infty$ must be contained in some $\mathbb{R}^n$ I'd like to prove that a compact subset of $\mathbb{R}^\infty$ must be contained in some finite dimensional space $\mathbb{R}^n$. Here, $\mathbb{R}^\infty$ is the set of all eventually zero sequences of reals, and the topology on $\mathbb{R}^\infty$ is defined as: $U\subset\mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n$ is open in each $\mathbb{R}^n$. I see that here we can show every convergent sequence in $\mathbb{R}^\infty$ (and its limit) must lie in some $\mathbb{R}^n$.
So I guess to prove the statement, we can suppose for a contradiction that a compact set $K$ is not contained in any $\mathbb{R}^n$. Then there is a sequence in $K$ that does not lie in any $\mathbb{R}^n$. So by the above, this sequence does not converge. Does this reach a contradiction? (Does compactness imply sequential compactness in $\mathbb{R}^\infty$?)
 A: Given $x \in \mathbb{R}^{\infty}$, let $m(x)$ be the smallest $m$ for which $x \in \mathbb{R}^m$, i.e., $m(x)=\min\{i \mid \pi_i(x)=0\}$. Let's call such $m(x)$ the dimension where $x$ lives. We then have the following intermediate lemma:

Lemma: If $A\subset \mathbb{R}^{\infty}$ has no two points living in the same dimension, then $A$ is closed and discrete.

Proof: The hypothesis tells us that $A \cap \mathbb{R}^n$ is finite for every $n$, hence closed. Since this holds for every $n$, we have that $A$ is closed. 
Note now that if $A$ has no two points living in the same dimension, then the same is true for $A \backslash \{x\}$ for any $x \in A$, and therefore $A \backslash \{x\}$ is also closed. Since $A$ is already closed, it follows that $A \backslash \{x\}$ is closed with respect to $A$, and thus $\{x\}$ is clopen in $A$. Thus, $A$ is discrete. $\blacksquare$
Now, take a compact $K$. If $K$ is not contained in some $\mathbb{R}^n$, then there are infinitely many $x_i \in K$ such that all $x_i$ live in different dimensions. The set $A:=\{x_i\}$ then satisfies the hypothesis of the previous lemma, which tells us that it must be closed and discrete. Since it is a closed set inside a compact one, it is itself compact. Being also discrete, it is finite, a contradiction.

The above is an adaptation of the proof that if $X$ is a CW-complex then every compact set is contained in a finite union of open cells.
A: Use contrapositive. Sketch: 
Suppose a subset $A \subseteq \mathbb{R}^{\infty}$ is contained in infinitely many $R_i$. Then construct an open cover in which each open set corresponds to one copy of $\mathbb{R}$. Then any finite subcover will not cover all of $A$.

Suppose $A \subseteq \mathbb{R}^{\infty}$ is a subset s.t. the intersection with infinitely many copies of $\mathbb{R}$ is non empty i.e. one which is not contained in $\mathbb{R}^n$ for some $n \in \mathbb{N}$. Then write $\mathbb{R}^{\infty} = \prod_{i \in \mathbb{N}} \mathbb{R}_i$ and denote $A_i := \pi_i(A)$. Then define $B_i \subseteq \mathbb{R}^{\infty}$ s.t. 
$$\pi_j(B_i) \begin{cases} = \emptyset, & i \neq j \\
\supseteq A_i, ~ \text{open} & i = j
\end{cases}
$$ 
Each $B_i$ is open in $\mathbb{R}^{\infty}$ since $B_i \cap \mathbb{R}^n$ is either empty or open for any $n \in \mathbb{N}$. Also, $B_i$ covers $A_i$ for every $i \in \mathbb{N}$. 
Thus $A$ has an open cover $A \subseteq \cup_{i \in \mathbb{R}} B_i$, however, any finite subcover does not cover $A$ since $A \cap \mathbb{R}_i \neq \emptyset$ for infinitely many $i \in \mathbb{N}$. Hence $A$ is not compact. 
A: Not a complete answer, but some "out loud thinking along", hopefully helpful:
$\mathbb{R}^n$ (as a subset of $\mathbb{R}^\infty$) is the set $$\{(x_i)_i \in \mathbb{R}^\infty: \forall i>n: x_i=0\}$$
and this has the the standard topology as a copy of $\mathbb{R}^n$ by the obvious homeomorphism  that just keeps the first $n$ coordinates of each such sequence.
This way we have that the $\mathbb{R}^n$ form an increasing family of subsets of $\mathbb{R}^\infty$, whose union is the whole space. This is the usual interpretation of this space AFAIK.
Consider for a fixed $m$ the set $O_m=\{(x_i)_i \in \mathbb{R}^\infty: x_m \neq 0\}$. If $n < m$ then $\mathbb{R}^n \cap O_m =\emptyset$ and if $n \ge m$ we have $\mathbb{R}^n \cap O_m = \pi_m^{-1}[\mathbb{R}\setminus\{0\}]$ and in both cases $O_m$ is open in $\mathbb{R}^n$, so all $O_m$ are open subsets of $\mathbb{R}^\infty$ in the given (inductive limit) topology. 
So if $K$ is compact then if $0 \notin K$, then the $O_m$, $m \in \mathbb{N}$, form an open cover of $K$ and finitely many of them must cover $K$, which means that $K$ is a subset of $\mathbb{R}^n$ for $n$ being the maximal of the finitely many $m$ used. If $0 \in K$, I don't as yet see a valid argument.. 
