Prove that the space of infinite sequences of real numbers is infinite dimensional I understand basically how to show this, but how would one prove it? 
To show this, we know a basis for this vector space would be: 
E1 = (1, 0, 0,...)
E2 = (0, 1, 0,...)
E3 = (0, 0, 1,...)
.
.
E_infinity 
Which means the dimension of the space is infinite. Is this sufficient? 
 A: As the comments indicate, the set you have chosen is an infinite linearly independent set, but it is not quite a "basis", by definition. (That is, it is not a "Hamel basis").
It is a theorem that in any finite dimensional space, the number of elements in a linearly independent set is at most the dimension of the space.  Thus, in finding an infinite linearly independent subset, you have shown that the space cannot be finite dimensional.
To further address your intuition: the span of your set is precisely the set of all sequences that end in infinitely many zeros.  Note that the definition of the "span" of a set of vectors considers only linear combinations of finitely many elements at a time.  In fact, we cannot guarantee that any infinite sums will "make sense" without imposing further structure on a vector space (such as a topology).
That being said, your set is what is commonly referred to as a "Schauder basis" of some sequence spaces.  That's right: there are different spaces of infinite sequences of real numbers, each of which carries a different notion of "distance" between two sequences.  I would invite you to consider what exactly it should mean for an infinite sum of sequences to "converge".
A: We will try to prove this by contradiction. First let us denote the space containing all the sequences by $V$. Let us first assume that it is finite dimensional and denote its dimension by $n$.


*

*Using the fact that any list of linearly independent vectors of length equal 
to $\dim(V)$ is a basis

*So we choose a set of sequences $T_1,T_2,T_3 \dots T_n$ such that sequence $T_1$ has all but 1st term zero, $T_2$ has all but 2nd term zero.

*Consider $T(n+1)$. We can show that it cannot be represented as linear combination of the basis we chose.



Hence by contradiction we have proved our statement
