Vector space that can be made into a Banach space but not a Hilbert space

No. We have the following result.

Proposition. Let $X$ be a real or complex vector space. Then $X$ can be furnished with a complete norm if and only if there exists an inner-product on $X$ whose corresponding norm is complete.

For a topological space $X$, denote by $d(X)$ the density of $X$, that is, the minimal cardinality of a dense set in $X$. We denote the cardinality of the continuum by $\mathfrak{c}$. Certainly, $\mathfrak{c}=\mathfrak{c}^{\aleph_0}$, which we shall need.

Proof. It is enough to prove the implication $(\Rightarrow$). Let $X$ be a Banach space. Without loss of generality $X$ we may suppose that $X$ is infinite-dimensional. We split the proof into two cases.

Case where $d(X)\leqslant \mathfrak{c}$.

We know that the cardinality $b(X)$ of any Hamel basis of an infinite-dimensional Banach space is at least $\mathfrak{c}$. Thus,

$$\mathfrak{c}\leqslant b(X)\leqslant |X|\leqslant d(X)^{\aleph_0}\leqslant \mathfrak{c}^{\aleph_0}=\mathfrak{c},$$

which yields $b(X)=\mathfrak{c}$^[a]. This means that $X$ is isomorphic as a vector space to the (separable!) Hilbert space $\ell_2$, so one may use any algebraic isomorphism between $X$ and $\ell_2$ to define a complete, inner-product norm on $X$.

Case where $d(X)> \mathfrak{c}$.

For Banach spaces $X$ with $d(X)> \mathfrak{c}$, the cardinality of $X$ is the same as $b(X)$. We then have $$b(X)=|X|=|\ell_2(d(X))|=b(\ell_2(d(X))),$$ so one may use any algebraic isomorphism between $X$ and the Hilbert space $\ell_2(d(X))$ to define a complete, inner-product norm on $X$. $\square$

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[a]: Actually one has the equality $|X|=d(X)^{\aleph_0}$ for every infinite-dimensional Banach space but we do not need it here.