Is there a vector space that cannot be an inner product space?

I'm assuming the ground field is ${\mathbb R}$ or ${\mathbb C}$, because otherwise it's not clear what an "inner product space" is.

Now any vector space $X$ over ${\mathbb R}$ or ${\mathbb C}$ has a so-called Hamel basis. This is a family $(e_\iota)_{\iota\in I}$ of vectors $e_\iota\in X$ such that any $x\in X$ can be written uniquely in the form $x=\sum_{\iota\in I} \xi_\iota\ e_\iota$, where only finitely many $\xi_\iota$ are $\ne 0$. Unfortunately you need the axiom of choice to obtain such a basis, if $X$ is not finitely generated.

Defining $\langle x, y\rangle :=\sum_{\iota\in I} \xi_\iota\ \bar\eta_\iota$ gives a bilinear "scalar product" on $X$ such that $\langle x, x\rangle>0$ for any $x\ne0$. Note that in computing $\langle x,y\rangle$ no question of convergence arises.

It follows that $\langle\ ,\ \rangle$ is an inner product on $X$, and adopting the norm $\|x\|^2:=\langle x,x\rangle$ turns $X$ into a metric space in the usual way.

How about vector spaces over finite fields? Finite fields don't have an ordered subfield, and thus one cannot meaningfully define a positive-definite inner product on vector spaces over them.

Christian Blatter's answer shows that, assuming the axiom of choice, every vector space can be equipped with an inner product.

Without the axiom of choice, this can fail. As I show in Inner product on $C(\mathbb R)$, it is consistent with ZF+DC that the vector space $C(\mathbb{R})$ of continuous functions on $\mathbb{R}$ does not admit any inner product, nor even any norm.

The idea is that $C(\mathbb{R})$ already has a "nice" topology which is not compatible with an inner product, and under appropriate axioms consistent with ZF+DC, there are "automatic continuity" results saying that it cannot have any other "nice" topology.