Set of infinite tuples is closed, bounded, non-compact 
Let $\mathbb{R}^{\infty}$ be the set of all "infinite-tuples" $x=(x_1,x_2,\ldots)$ of real numbers that end in an infinite string of $0$'s. Define an inner product on $\mathbb{R}^{\infty}$ by the rule $(x,y)=\sum x_iy_i$. Let $\|x-y\|$ be the corresponding metric on $\mathbb{R}^{\infty}$. Define $e_i=(0,\ldots,0,1,0,\ldots,0,\ldots)$, where $1$ appears in the $i$th place. Let $X$ be the set of all the points $e_i$. Show that $X$ is closed, bounded, and non-compact.

$X$ is bounded: Clear, since $\|e_i\|=1$ for all $i$.
$X$ is closed: Suppose that $y=(y_1,y_2,\ldots)$ is a limit point of $X$, and let $y_n=y_{n+1}=\ldots=0$. Then for $i\geq n$, we have $$\|e_i-y\| = \sqrt{\sum_j ((e_i)_j-y_j)^2}\geq \sqrt{((e_i)_i-y_i)^2} = 1.$$ Let $d=\min(\|e_1-y\|,\|e_2-y\|,\ldots,\|e_{n-1}-y\|,0.5)$. If $d>0$, the ball of radius $d/2$ centered at $y$ doesn't contain any element of $X$, and so $y$ is not a limit point. So $d=0$, and $y=e_i$ for some $i$. Hence $X$ is closed.
$X$ is not compact: We must show that there exists an open collection covering $\{A_{\alpha}\}$ of $X$ for which no finite subcollection covers $X$. What collection might we choose here?

Hint: $(e_n)_{n=1}^\infty $ doesn't have a convergent subsequence.
