Finding a basis for a set of tuples 
Let $R^\infty$  be the subspace of U={${f:N→R}$} of “infinite tuples” which consist of all $f\in U$ such that f(i) = 0 for all but finitely many values of i. Define an inner product on  $R^\infty$ by (x, y) $=\sum_{i=1}^\infty xiyi$. Let $∥x − y∥$ be the metric on $R^\infty$. Find the basis of $R^\infty$ and prove the set of basis is closed, bounded, and noncompact.

This is a finite sum for the inner product according to the above definition of the subspace, since all but finitely many terms vanish. For the first part, I defined $e_i = (0,··· ,0,1,0,··· ,)$, where 1 appears on the ith slot and I think ei form a basis for $R^\infty.$ I am not sure how to show this set is the base though.
For the second part the set would be {${e_i : i ∈ N}$} but I thought a set is compact if it is closed and bounded but how can it be noncompact if it is closed and bounded?
Finally, what does it mean by a set of infinite tuples?
 A: By definition, if $f\in R^\infty$ you have 
$$
f=\sum_{j=1}^m f_j\,e_j. 
$$
So the $\{e_j\}$ span $R^\infty$. And they are linearly independent: if $f=\sum_{j=1}^m a_j e_i=0$, then $0=f(k)=a_k$ for all $k$. 
It is very not true that "closed + bounded = compact". It is always true in finite dimension, and it not true in infinite dimension. The set $\{e_i\}$ is clearly non-compact, since it doesn't admit a convergent subsequence: you have $\|e_k-e_j\|=\sqrt2$ for all $k,j$. 
Finally, all the above is talking about sequences. A sequence of real numbers is nothing other than a function $f:\mathbb N\to\mathbb R$. The usual sets in this context are 


*

*$\ell^\infty(\mathbb N)$: the bounded sequences

*$c_0$: the sequences that converge to $0$

*$\ell^1(\mathbb N)$: the sequences $f$ with $\sum_j|f(j)|<\infty$

*$\ell^2(\mathbb N)$: the sequences $f$ with $\sum_j|f(j)|^2<\infty$

*etc. 
The notation $R^\infty$ is unusual, but (even though it is ever present) there is no universally accepted notation for it
A: Firstly, you've got a slight notational weirdness: $e_i$ should really be a function. It's clear what function you mean, but mixing notations is odd. Instead, define $$e_i(n) = \cases{1, & n = i\\ 0, & \text{otherwise}}$$
You can check that it is a basis directly: for any $f \in U$, we have $f = \sum_i f(i)e_i$, since for all $n$, we have $f(n) = f(n)e_n(n) = \sum_i f(i)e_i(n)$ (where in the first equality, we're multiplying by $1$, and in the second, we're adding $0$). 

I thought a set is compact if it is closed and bounded but how can it be noncompact if it is closed and bounded?

This is true for subsets of $\mathbb{R}^n$. Your set (which I'll call $X$) clearly is not a subset of $\mathbb{R}^n$. I rather suspect that this exercise is here purely to point out that said equivalence does not hold outside of $\mathbb{R}^n$.
The bound on $X$ is simple: $\|e_i\| = \sum_j e_i(j)^2 = e_i(i)^2 = 1$ for all $i$. 
To see that $X$ is closed, first show that $\|e_i - e_j\| = 2$ for all $i \neq j$, and so the only convergent sequences in $X$ are the eventually constant sequences.
To see that $X$ is not compact, define $A_i$ to be the open $1$-ball around $e_i$ in $U$ for all $i$. This gives, after intersecting with $X$, an open cover of $X$ (clearly each $A_i$ is open in $U$, so its intersection with $X$ is open in $X$, and clearly it's a cover, as each $e_i$ lies in $A_i$. But $\{A_i\}$ has no finite subcover - indeed, by what we did to show that $X$ is closed, we see that the $A_i$ are disjoint, and so $\{A_i\}$ has no proper subcovers at all, so certainly has no finite ones. Thus, $X$ is not compact. 
