TITLE: Proofs involving a set of infinite dimensional vectors
QUESTION [5 upvotes]: Given: 
A = $\{  \vec{v} = (v_{1}, ....v_{k}, ...) \in \mathbb{R}^{\infty} | \sum_{i=1}^\infty v_i^2 \text{converges} \}$
Prove that the set is a subspace of $\mathbb{R}^\infty$ and:
< $\vec{v}, \vec{u} > = \sum_{i=1}^\infty v_iu_i$ defines an inner product on A.
I am really struggling with this problem, particularly showing closure under addition to show the set is a subspace, and proving positive definiteness of the inner product. I have been given the hint by a professor "use the Cauchy-Schwarz inequality" but I lack confidence in this advice, it is my understanding that a well defined inner product is a pre-condition for using Cauchy-Schwarz inequality. 
This is associated with an intro Linear Algebra take home exam, I have found a lot of information that leads me to believe this is regarding an $\mathscr{l}^2$ - Hilbert space, a concept that is not actually in my textbook.
How to go about this proof? Ideally in a way that someone with 1 semester of linear algebra could understand.

REPLY [2 votes]: Let  $x,y \in A$ where $$x=(x_1,x_2....)$$ $$y=(y_1,y_2....)$$
Then $x+y=(x_1+y_1,x_2+y_2.....)$ and 
$\sum_{n=1}^{\infty} (x_n+y_n)^2=\sum_{n=1}^{\infty}x_n^2+2x_ny_n+y_n^2 \leqslant 
\sum_{n=1}^{\infty}2x_n^2+2y_n^2=2\sum_{n=1}^{\infty}x_n^2+2\sum_{n=1}^{\infty}y_n^2 < \infty$  thus $x+y \in A$(i used the fact that $x^2+y^2 \geqslant 2xy \Longleftrightarrow(x-y)^2 \geqslant 0$)
If  $a \in \mathbb{R}$ and $x=(x_1,x_2...) \in A$ then $\sum_{n=1}^{\infty}a^2x_n^2=a^2\sum_{n=1}^{\infty}x_n^2 < \infty$ thus $ax \in A$.
We proved that $A$ is a subbase of $\mathbb{R}^{\infty}$
Now $<x,x> $ where $x \in A$ is non negative because it is a sum of squares.
It is easy to prove the other axioms of inner product