I'm having trouble answering the question on the title due to the fact that I reach opposite conclusions based on 2 different reasonings. I'd be grateful if you could have a look at them and tell me what's wrong with one (or both) of them:
$\textbf{Reasoning 1}$: The statement in the title is true
Let $V$ be a vector space with some basis $B_V$ and let $G$ be the matrix associated to the scalar product in that given basis. Also, let $G_c$ be the matrix associated to the scalar product in the canonical basis, $B_c$, which we know is diagonal because $<\vec{e}_i, \vec{e}_j> = 0$ if $i \neq j$.
Since the matrixes $G$ and $G_c$ are related according to $G = P^TG_cP$ (and $G_c$ will always exist because the scalar product is always defined for the canonical basis), we know $G$ and $G_c$ are congruent matrixes, which implies they are also equivalent.
By definition, a matrix $A$ is diagonalizable if we can find a matrix $P$ such that $A = P^{-1}DP$, where $D$ is the diagonalized matrix. That is to say, a matrix will be diagonalizable if we can find a diagonal matrix to which it is equivalent.
Since we know our scalar product matrix $G$ will always be congruent with and therefore equivalent to a diagonal matrix $G_c$, then a scalar product matrix can always be diagonalized $\blacksquare$
$\textbf{Reasoning 2}$: The statement in the title is false
We know G will always be a square matrix, and, according to Reasoning 1, we will always be able to diagonalize G. Furthermore, since our basis $B_V$ can be whichever we like as long as it generates the vector space $V$, the scalar product matrix $G$ can take any value.
Since $G$ can be any square matrix and $G$ is always diagonalizable, every square matrix is diagonalizable, which is evidently not true.