There does not exist a fibre functor over $\mathbb{R}$. Indeed, the abelian variety over $\mathbb{F}_p$ corresponding to the Weil number $p^{1/2}$ has dimension 2. Its endomorphism algebra is a division algebra $E$ of degree $4$ over the field $F=\mathbb{Q}[p^{1/2}]$ that is ramified only at the two infinite primes of $F$. Clearly $E\otimes\mathbb{R}$ doesn't act on a $4$ dimensional $\mathbb{R}$ vector space, and so there is no Weil cohomology over $\mathbb{R}$ (there is over $\mathbb{Q}_{\ell}$ for all $\ell$).
In common English, the statement
I do not run every day.means
There are some days on which I do not run.Similarly,
$a(n)$ is not zero for all n.means
There are some $n$ for which $a(n)$ is not zero.For me at least, the following statements are equivalent:
$a(n)$ is not zero for all $n$.Lenstra uses the last to mean "no $a(n)$ is zero", which I certainly find confusing. In fact, I had to read his proof to make sure I was (miss)interpreting it correctly. This may be a Dutch problem. Oort (arXiv:math.AG/0701479v1, p81) writes that a simple algebra "does not split at every real place of $L$" when he means that it "splits at no real place of $L$". This is clearly wrong.
$a(n)$ is not equal to zero for all $n$.
$a(n)\neq0$ for all $n$.