Can every real vector space be turned into a Euclidean Vector Space? Here's what I'm reading right now:

So, the question that I have relating to this is if it needs to be proven that an inner product can be defined on every real vector space and if there are infinitely many inner products that can be defined on each real vector space, with the exception of $\{0\}$. If there is a proof for this, how would it go? I'm somewhat confused by Dr Klaus Janich's wording because it seems like this is something that can be proved.
Also, the way that it has been worded kind of gives me the impression that one can prove this by taking a real vector space and showing that there is a way to construct a set of inner products on that vector space. Then, you'd have to show that that set has infinitely many elements. Would this be a correct way of looking at it or am I just getting this entirely wrong?
 A: Suppose $V$ is a real vector space with inner product $(\cdot, \cdot)$. Then for any $c >0$ define inner product $\langle \cdot, \cdot \rangle$ by $\langle x,y\rangle = c(x,y)$. Show that this defines a perfectly good inner product.
Thus if we have inner product (not on the zero space, the only space for which the inner product is just the constant function to 0), we have infinitely many, just given by scaling as above.
Now what's left to be shown is any real vector space admits an inner product. Let $V$ a real vector space. By a standard result, every real vector space has a basis (note this requires the Axiom of Choice in the infinite case and this is necessary for this proof; this is really a technical detail you don't have to worry about i you don't want to/don't understand though). Let $\{b_i\}_{i \in I}$ be a basis for $V$. Then define an inner product $(x,y) = \sum_i x_i y_i$ where $x_i, y_i$ denote the coefficient on $b_i$ of $x$ (resp. $y$) represented in the basis. Note that only finitely many of the $x_i$ and finitely many of the $y_i$ will be nonzero, thus that sum makes sense and is finite. This is from the definition of a basis. You can check this really defines an inner product.
