Is every Banach space densely embedded in a Hilbert space?

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.

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 $$\|x\|_2^2 = \sum_n 2^{-n}f_n(x)^2 \leq \sum_n 2^{-n} \|x\|^2=\|x\|^2$$

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).