Is every Banach space densely embedded in a Hilbert space? Can every Banach space be densely embedded in a Hilbert space? This is clear if the Banach space is actually a Hilbert space, but much can you relax this?
If the embedding exists, is the target Hilbert space unique?

I will show that the answer is yes, for any separable Banach space $B$. The non-separable case is covered by George Lowther above.
So let $B$ be a separable Banach space, and consider some countable dense subset $\{x_n\}_{n\in \Bbb N} \subset B$. For each $n$ we apply the Hahn Banach theorem to find some $f_n \in B^*$ such that $f_n(x_n)=\|x_n\|$ and $\|f_n\|=1$.
We then define an inner product $\langle \cdot, \cdot \rangle_2$ on $B$ by setting $$\langle x,y \rangle_2 = \sum_{n=1}^{\infty} 2^{-n} f_n(x)f_n(y)$$ and we let $\|\cdot\|_2$ be the asociated norm.
To check that this is actually an inner product, the only nontrivial property we need to check is nondegeneracy (which means that $\|x\|_2=0 \implies x=0$). Define $g:=\sup_n f_n$. Note that $|g(x)| \leq \|x\|$ since $\|f_n\|= 1$ for all $n$. Also notice that $g$ is continuous, because for any fixed $N$ we have that $f_N(x)-\sup_nf_n(y) \leq f_N(x)-f_N(y)$ so taking the sup over all $N$ on both sides gives that $g(x)-g(y) \leq g(x-y) \leq \|x-y\|$. Symmetrically we can switch $x$ and $y$ so this actually shows that $|g(x)-g(y)| \leq \|x-y\|$, thus proving continuity. Also $g(x_n) \geq f_n(x_n)=\|x_n\|$ and thus $g(x_n)=\|x_n\|$ for all $n$. By density of the $x_n$ and continuity of $g$ it follows that $g(x) = \|x\|$ for all $x$. Hence $\|x\|=\sup_n f_n(x)$, so that if $\|x\|_2=0$ then $f_n(x)=0$ for all $n$ and thus $x=0$.
We denote by $B'$ the completion of $B$ with respect to $\|\cdot\|_2$. Then the inclusion map $i:B \to B'$ is a continuous embedding, because  $$\|x\|_2^2 = \sum_n 2^{-n}f_n(x)^2 \leq \sum_n 2^{-n} \|x\|^2=\|x\|^2$$
