Can all real/complex vector spaces be equipped with a Hilbert space structure? Let $X$ be a vector space over $\mathbb K \in \{\mathbb R, \mathbb C\}$. 
Does there exists a pairing $X \times X \rightarrow \mathbb K$ that induces a Hilbert space structure on $X$?
I have been thinking that one may choose an arbitrary basis of $X$, which exists by the axiom of choice, and declare it as the orthonormal basis of the Hilbert space, but then limits of series might become unhandy.
 A: No. If $X$ is a space of countably infinite dimension, then the Baire category theorem implies that $X$ can't be given the structure of a Hilbert space (or even a Banach space). Your idea fails since, in an infinite-dimensional Hilbert space, an orthonormal basis is never an algebraic (Hamel) basis. 
A: You may be interested in the following paper:

Lorenz Halbeisen, and Norbert Hungerbühler. The cardinality of Hamel bases of Banach spaces, East-West Journal of Mathematics, 2, (2000) 153-159.

There, Lorenz and Norbert prove a few results about the size of Hamel bases of arbitrary infinite dimensional Banach spaces. In particular, they show:

Lemma 3.4. If $K\subseteq\mathbb C$ is a field and $E$ is an infinite dimensional Banach space over $K$, then every Hamel basis of $E$ has at least cardinality $\mathfrak c=|\mathbb C|$.
Theorem 3.5. Under the above assumptions, the size of any Hamel basis of $E$ is $|E|$.

Among other things, this implies a negative answer to your question, and addresses Miha Habič's comment to the other answer.
The argument is short, and uses a key set theoretic fact, namely, that there are independent families of size $\mathfrak c$. Recall that an independent family is a collection $I$ of infinite subsets of $\mathbb N$ with the property that for any distinct $A_1,\dots,A_m,B_1,\dots,B_n$ sets in $I$, we have that
$$ \bigcap_{i=1}^m A_i\setminus\bigcup_{j=1}^n B_j\quad\mbox{ is infinite.} $$
(Lorenz has a few papers on set theoretic issues in Banach spaces, that can be downloaded from his page.)
