True or False: Every finite dimensional vector space can made into an inner product space with the same dimension. Every finite dimensional vector space can made into an inner product space with the
same dimension.
 A: The definitions of inner product space that I have seen always require that vectors have a nonnegative inner product with themselves. Since inner products live in the base field, this requirement can only be meaningful if that field is of characteristic $0$. So your statement would be false over fields of prime characteristic. Also over fields larger than $\Bbb C$, like $\Bbb C(X)$, I believe it would be hard to arrange that the inner products of vectors with themselves lie in an ordered subfield like $\Bbb R$.
If (as is usual) you consider inner product spaces only over the fields $k=\Bbb R$ or $k=\Bbb C$, then indeed every finite dimensional vector space can be made into an inner product space, by transport via a vector space isomorphism with $k^n$ of the standard inner product on the latter.
A: I think it depends on what field you are using for your vector space.  If it is $\mathbf{R}$ or $\mathbb{C}$, the answer is definitely "yes" (see the comments, which are correct).  I am pretty sure it is "yes" if your field is a subfield of $\mathbb{C}$ that is closed (the word "stable" also seems to be standard) under complex conjugation, such as the algebraic numbers. Otherwise, e.g. if your field is $\mathbf{F}_2$, I don't know.  I consulted Wikipedia's article on inner product spaces and they only dealt with the case where the field was the reals or the complex numbers.
EDIT: Marc's answer is better than mine.  See his comment regaring subfields of $\mathbf{C}$.  Some such subfields are not stable under complex conjugation and cannot be used as a field for an inner product space. I am pretty sure that if $\mathbb{K}$ is any ordered field (which may or may not be a subfield of $\mathbb{R}$) you can use it as the field for an inner product space. 
