Infinite Sphere is connected but not compact $\mathbb{R}^{\infty}$: Space of sequences $(x_i)_{i=1}^{\infty}$ of real numbers such that at most finitrly many of $x_i$'s are nonzero
Embed $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ via $(x_1,x_2,\ldots,x_n) \to (x_1,x_2,\ldots,x_n,0)$. Then $\mathbb{R}^{\infty}=\cup_n\mathbb{R}^n$
Define a topology on $\mathbb{R}^{\infty}$ by declaring that a set $C \subset \mathbb{R}^{\infty}$ is closed iff $C \cap \mathbb{R}^n$ is closed in $\mathbb{R^n}$ for every $n$. 
Let $S^\infty=\cup_{n}S^n$. Show that $S^\infty$ is connected but not compact. 
Suppose that $S^{\infty}=C \cup D$ where $C$ and $D$ are two disjoint closed sets in $\mathbb{R}^{\infty}$. Then $S^m=(S^m\cap C)\cup (S^m \cap D)$. Since $C$ and $D$ are closed in $\mathbb{R^{\infty}}$, $C\cap \mathbb{R}^m$ and $D \cap \mathbb{R}^m$ are closed in $\mathbb{R}^m$. 
So $S^m=(S^m \cap C \cap \mathbb{R}^m) \cup (S^m \cap D \cap \mathbb{R^m})$ can be written as a disjoint union of closed sets implying that $S^m$ is not connected which is a contradiction. 
Let $A_n=B_n(0,r)$ for each $n \in \mathbb{N}$ for $r \gt 1$ .Then $A_n$ is open and $S^{\infty}=\cup_n A_n$. But this cover doesn't have a finite subcover. 
Is this alright?
Thanks for the help!!
 A: Here is another fun proof. Consider the map
$$ f \colon S^{\infty} \to \mathbb{R} $$
$$ (x_0,x_1,\ldots) \mapsto \sum n x_n $$
This map is well defined since any element of $S^{\infty}$ has only a finite number of nonzero $x_i$. And it is continuous because the restriction to each $S^n$ is continuous. The images of the points $(0,\ldots,0,1,0, \ldots)$ and $(0,\ldots,0,-1,0, \ldots)$ are precisely the integers and so all the integers belong to the image. Since the map is continuous, it must be surjective. But $\mathbb{R}$ is not compact, so $S^{\infty}$ can not be compact. 
A: Let $B=\{e_n:n\ge1\}$ where $e_n=(0,\ldots,0,1,0,\ldots)$, the $1$ being in the $n^\text{th}$ position. Then $B\cap\mathbb R^n=\{e_k:1\le k\le n\}$ which is closed, hence $B$ is closed. Clearly $B\subset S^\infty$, so it suffices to show $B$ is not compact. Let $U_n=\{x=(x_1,x_2,\ldots):|x_n-1|<\frac12\}$. Then $U_n^c=\{x:|x_n-1|\ge\frac12\}$ which is clearly closed, so $U_n$ is open. Moreover, $e_n\in U_m$ if and only if $n=m$, so $\{U_n\}$ is a cover of $B$ with no finite subcover.
