Proof, [Theorem III.7.4, SW99]. For each $n \in \natp$, identify $E_{n}$ as a subspace of $\bigoplus_{n = 1}^{\infty} E_{n}$. Let $F$ be a Banach space and $T \in L(\bigoplus_{n = 1}^{\infty} E_{n}; F)$. For each $n \in \natp$, $E_{n}$ is a nuclear space, so $T|_{E_n}: E_{n} \to F$ is a nuclear operator, and there exists $\bracsn{\phi_{n, k}}_{k = 1}^{\infty} \subset E_{n}^{*}$ equicontinuous, $\bracsn{y_{n, k}}_{k = 1}^{\infty} \subset B_{F}(0, 1)$, and $\bracsn{\lambda_{n, k}}_{k = 1}^{\infty} \subset K$ such that $\sum_{k \in \natp}|\lambda_{n, k}| \le 2^{-n}$ and

for all $x \in E_{n}$. Thus for any $x \in \bigoplus_{n = 1}^{\infty} E_{n}$,

where $\sum_{n \in \natp}\sum_{k \in \natp}|\lambda_{n, k}| \le \sum_{n \in \natp}2^{-n}< \infty$ and $\bracsn{y_{n, k}|n, k \in \natp}\subset B_{F}(0, 1)$.

Finally, for each $n \in \natp$, let $U_{n} = \bigcap_{k \in \natp}\phi_{n, k}^{-1}(B_{K}(0, 1))$, then $U_{n} \in \cn_{E_n}(0)$ by equicontinuity of $\bracsn{\phi_{n, k}}_{k = 1}^{\infty} \subset E_{n}^{*}$. Let $U = \aconv(\bigcup_{n \in \natp}U_{n})$, then $U \in \cn_{\bigoplus_{n = 1}^\infty E_n}(0)$ and $U \subset \bigcap_{n \in \natp}\bigcap_{k \in\natp}(\phi_{n, k}\circ \pi_{n})^{-1}(B_{K}(0, 1))$. Hence $\bracsn{\phi_{n, k} \circ \pi_n|n, k \in \natp}$ is equicontinuous, and $T$ is a nuclear operator.$\square$