$\newcommand{\mc}{\mathcal}$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.

Edit: As Tomek points out, if $E$ is reflexive and $\mc U$ is countably complete, then $(E)_{\mc U} = E$ canonically. Here's a proof (which I hadn't realised before). That $\mc U$ is countably complete is equivalent to the property that if $(A_n)_{n=1}^\infty$ is a sequence in $\mc U$ then also $\cap_n A_n \in \mc U$. Let $\mc U$ be on a set $I$, and let $A_n\subseteq I$ be any sequence of subsets which cover $I$, so $\cup_n A_n = I$. We claim that then some $A_n\in\mc U$. For if not, $I\setminus A_n\in\mc U$ for all $n$ (as $\mc U$ is an ultrafilter) and so $\cap_n (I\setminus A_n) = \emptyset\in\mc U$, contradiction.

Let $(x_n)$ be a dense sequence in $E$, and let $(y_i)\in (E)_{\mc U}$. Consider the sets $$ A_{n,m} = \{ i : \|y_i - x_n\| < 1/m \}. $$ As $(x_n)$ is dense, for any fixed $i$ and $m$ there is some $n$ with $\|y_i-x_n\|<1/m$. So $(A_{n,m})$ covers $I$, and so there is some $A_{n,m}\in\mc U$. Fix this $n$ and consider $A_{n,k}\subseteq A_{n,m}$ for $k\geq m$. Repeating the argument finds that there is an increasing sequence $m \leq k_1 < k_2 < k_3 < \cdots$ with $A_{n,k_i}\in\mc U$ for each $i$. Hence $$ \bigcap_i A_{n,k_i} = \{ i : y_i=x_n \} \in\mc U $$ and so $(y_i) = x_n\in E$ in $(E)_{\mc U}$.

$\newcommand{\mc}{\mathcal}$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.

Edit: As Tomek points out, if $E$ is separable (with no other condition) and $\mc U$ is countably complete, then $(E)_{\mc U} = E$ canonically. Here's a proof (which I hadn't realised before). That $\mc U$ is countably complete is equivalent to the property that if $(A_n)_{n=1}^\infty$ is a sequence in $\mc U$ then also $\cap_n A_n \in \mc U$. Let $\mc U$ be on a set $I$, and let $A_n\subseteq I$ be any sequence of subsets which cover $I$, so $\cup_n A_n = I$. We claim that then some $A_n\in\mc U$. For if not, $I\setminus A_n\in\mc U$ for all $n$ (as $\mc U$ is an ultrafilter) and so $\cap_n (I\setminus A_n) = \emptyset\in\mc U$, contradiction.

Let $(x_n)$ be a dense sequence in $E$, and let $(y_i)\in (E)_{\mc U}$. Consider the sets $$ A_{n,m} = \{ i : \|y_i - x_n\| < 1/m \}. $$ As $(x_n)$ is dense, for any fixed $i$ and $m$ there is some $n$ with $\|y_i-x_n\|<1/m$. So $(A_{n,m})$ covers $I$, and so there is some $A_{n,m}\in\mc U$. Fix this $n$ and consider $A_{n,k}\subseteq A_{n,m}$ for $k\geq m$. Repeating the argument finds that there is an increasing sequence $m \leq k_1 < k_2 < k_3 < \cdots$ with $A_{n,k_i}\in\mc U$ for each $i$. Hence $$ \bigcap_i A_{n,k_i} = \{ i : y_i=x_n \} \in\mc U $$ and so $(y_i) = x_n\in E$ in $(E)_{\mc U}$.

$\newcommand{\mc}{\mathcal}$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.

$\newcommand{\mc}{\mathcal}$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.

Edit: As Tomek points out, if $E$ is reflexive and $\mc U$ is countably complete, then $(E)_{\mc U} = E$ canonically. Here's a proof (which I hadn't realised before). That $\mc U$ is countably complete is equivalent to the property that if $(A_n)_{n=1}^\infty$ is a sequence in $\mc U$ then also $\cap_n A_n \in \mc U$. Let $\mc U$ be on a set $I$, and let $A_n\subseteq I$ be any sequence of subsets which cover $I$, so $\cup_n A_n = I$. We claim that then some $A_n\in\mc U$. For if not, $I\setminus A_n\in\mc U$ for all $n$ (as $\mc U$ is an ultrafilter) and so $\cap_n (I\setminus A_n) = \emptyset\in\mc U$, contradiction.

Let $(x_n)$ be a dense sequence in $E$, and let $(y_i)\in (E)_{\mc U}$. Consider the sets $$ A_{n,m} = \{ i : \|y_i - x_n\| < 1/m \}. $$ As $(x_n)$ is dense, for any fixed $i$ and $m$ there is some $n$ with $\|y_i-x_n\|<1/m$. So $(A_{n,m})$ covers $I$, and so there is some $A_{n,m}\in\mc U$. Fix this $n$ and consider $A_{n,k}\subseteq A_{n,m}$ for $k\geq m$. Repeating the argument finds that there is an increasing sequence $m \leq k_1 < k_2 < k_3 < \cdots$ with $A_{n,k_i}\in\mc U$ for each $i$. Hence $$ \bigcap_i A_{n,k_i} = \{ i : y_i=x_n \} \in\mc U $$ and so $(y_i) = x_n\in E$ in $(E)_{\mc U}$.