$\newcommand{\mc}{\mathcal}$The confusion seems to be over the following claim:

> **Claim:** Let $\mc U$ be a countably incomplete ultrafilter, and let $E$ be a Banach space.  If $(E)_{\mc U}$ is reflexive, then $E$ is super-reflexive.

To start with, we use the Eberlein-Smulian Theorem to observe that a Banach space $F$ is reflexive if and only if every _separable_ subspace of $F$ is reflexive.  [Question: Do we need Eberlein-Smulian to show this?]

Recall also that $\mc U$ being _countably incomplete_ means that there is a nested sequence of sets $A_1 \supseteq A_2 \supseteq \cdots$ in $\mc U$ with $\cap_i A_i = \emptyset$.  I think of this property as allowing us to embed sequential convergence into convergence along $\mc U$.

Finally, let us recall Theorem 6.3 in Heinrich's paper:

> If $F$ is a separable Banach space finitely representable in $E$ then $F$ embeds isometrically into $(E)_{\mc U}$ for any countably incomplete $\mc U$.

Suppose towards a contradiction that $E$ is not super-reflexive, so there is a non-reflexive $F$ finitely representable in $E$.  There is hence a separable subspace $F_0$ of $F$ which is not reflexive.  Clearly $F_0$ is still finitely representably in $E$, and so isometric to a subspace of $(E)_{\mc U}$.  Hence $(E)_{\mc U}$ is not reflexive, contrary to assumption.

We then proceed exactly as Jochen Glueck's comment.