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?
 A: 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  $$\|x\|_2^2 = \sum_n 2^{-n}f_n(x)^2 \leq \sum_n 2^{-n} \|x\|^2=\|x\|^2$$
A: Suppose that there exists an injective, bounded linear operator $T\colon X\to H$, where $H$ is a Hilbert space. Then $\|x\|_{\rm new}=\|x\|+\|Tx\|$ ($x\in X$) defines an equivalent, strictly convex norm on $X$.
Not every Banach spaces admits a strictly convex renorming -- just to mention $\ell_\infty(\omega_1)$ (Day) or $\ell_\infty / c_0$ (Bourgain).
A: We can show that $\ell_\infty(I)$ cannot be embedded into a Hilbert space, for an uncountable index set $I$.
Here, I am interpreting the question as Normal Human suggests - where the embedding is assumed to be linear and continuous. Suppose that $f\colon \ell_\infty(I)\to H$ is an embedding into Hilbert space $H$. By continuity, there exists a $K\in\mathbb{R}$ such that $\lVert f(x)\rVert\le K\lVert x\rVert$ for all $x\in\ell_\infty(I)$.
Given any sequence $x_1,\ldots,x_n\in H$, the identity
$$
\sum_{\epsilon_1,\ldots,\epsilon_n=\pm1}\left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2=2^n\left(\sum_{r=1}^n\lVert x_r\rVert^2\right)
$$
holds. This implies that there exists a sequence $\epsilon_r\in\{\pm1\}$ such that
$$
\left\lVert\sum_{r=1}^n\epsilon_rx_r\right\rVert^2\ge\sum_{r=1}^n\lVert x_r\rVert^2.
$$
Now, for each $i\in I$, let $e_i\in\ell_\infty(I)$ be defined by $(e_i)_j=0$ for $j\not=i$ and $(e_i)_i=1$. Also, for each $n\in\mathbb{Z}_{>0}$, let $S_n$ be the set of $i\in I$ such that $\lVert f(e_i)\rVert\ge1/n$. As $f$ is an embedding, we have $f(e_i)\not=0$, so $\bigcup_{n=1}^\infty S_n=I$ is uncountable. Hence, $S_n$ is infinite for some $n$. Then, for any $N > 0$, pick a sequence $i_1,\ldots,i_N$ of distinct elements of $S_n$. By what we showed above, there is a sequence $\epsilon_1,\ldots,\epsilon_N\in\{\pm1\}$ such that $x\equiv\sum_{r=1}^N\epsilon_re_{i_r}$ satisfies
$$
\lVert f(x)\rVert^2\ge\sum_{r=1}^N\lVert f(e_{i_r})\rVert^2\ge N/n^2.
$$
However, $\Vert x\rVert=1$, so
$$
\lVert f(x)\rVert\ge n^{-1}\sqrt{N}\lVert x\rVert.
$$
Choosing $N>K^2n^2$ gives a contradiction.
A: I could be wrong, but it seems you are asking simply "can we define an inner product on an arbitrary Banach space?", which is addressed in this question: Is there a vector space that cannot be an inner product space? 
Specifically, in the second answer in the question listed, by the AOC notice we can write down a Hamel basis $\{e_i\}$ for $X$, then define $(x,y)=\sum\overline{a_i}b_i$ for $x=\sum a_i e_i$ and $y=\sum b_i e_i$. This will give an inner product on $X$, and so $H=(X,\sqrt{(,)})$ will be a Hilbert space, and hence $T\colon X\hookrightarrow H$ will be such an embedding. This need not be unique since the the inner product depends on the choice of basis, and indeed as pointed out in the comments will not be isometric, and the norm will almost certainly be different.
I hope this helps.
