Is the space $\ell^\infty$ a Hilbert space? 
Is the space $\ell^\infty$ a Hilbert space?

I'm wondering if we can verify this using the parallelogram law? If $x = (1,0,0,\dots)$ and $y=(0,1,0,0, \dots)$ are both in $\ell^\infty$, then $$\|x+y\|^2_\infty = 1 \ne 2\|x\|^2_\infty + 2\|y\|^2_\infty = 4 $$
so this space does not satisfy the parallelogram law. Would this mean that space is not a Hilbert space if so why?
 A: This is a long comment that I rather put in the answer section.
The space $(\ell_\infty(\mathbb{N}),\|\;\|_\infty)$ as the OP indicated is not a Hilbert space in the sense that the norm $\|\;\|_\infty$ is not induced by an inner product (the parallelogram identity does not hold)
It is possible to define an inner product on  $\ell_\infty$ so that, with the induced norm $\|\;\|$, $(\ell_\infty,\|\;\|)$ is a subspace of a  Hilbert space $H$, for example $\langle \mathbf{x},\mathbf{y}\rangle =\sum_n a_nx_ny_n$, where $a_n>0$ and $\sum_na_n<\infty$. The Hilbert $H$ containing $\ell_\infty(\mathbb{N})$ is given by $H=\{\mathbf{x}\in\mathbb{R}^{\mathbb{N}}:\sum_na_n |\mathbf{x}(n)|^2<\infty\}$.
Notice that the inclusion map $\iota:(\ell_\infty,\|\;\|_\infty)\rightarrow (H,\|\;\|)$ is bounded:
$$\|\mathbf{x}\|\leq\Big(\sum_na_n\Big)\|\mathbf{x}\|_\infty$$
However the norms $\|\;\|$ and $\|\;\|_\infty$ are not equivalent  on $\ell_\infty(\mathbb{N})$, otherwise $\iota^{-1}:(\ell_\infty,\|;\|)\rightarrow(\ell_\infty,\|\;\|_\infty)$ would be a continuous linear operator and so there would be a constant $c>0$ such that $\|\mathbf{x}\|_\infty\leq c\|\mathbf{x}\|$.
This means $(\ell_\infty,\|\;\|_\infty)$ and $(\ell_\infty,\|;\|)$ would have the same Cauchy sequences. This is not the case however: the sequence $\mathbf{e}_m:n\mapsto\mathbb{1}_{\{1,\ldots,m\}}(n)$ is Cauchy in $(\ell_\infty,\|\;\|)$   but not in $(\ell_\infty,\|\;\|_\infty)$.
The question is whether any such norm $\|\;\|$ induced the same topology as that induced by $\|\;\|_\infty$.
The key point, I think, is to notice that under $\|\;\|_\infty$, $\ell_\infty$ is not separable. Under the inner product above, $\ell_\infty$ becomes separable.
The solution to the OP would follow if one proves

Suppose $\|\;\|$ is a norm on $\ell_\infty(\mathbb{N})$ that is induced by an inner product. Then $(\ell_\infty(\mathbb{N}), \|\;\|)$ is separable.

I have not checked the details, but It seems to be true.
