= Solution
Fix $z_0\in E$ and $\epsilon>0$. Choose $N$ so that $\sup_E|f_N-f|<\epsilon/3$, and then choose $\delta>0$ so that $|f_N(z)-f_N(z_0)|<\epsilon/3$ for $z\in E$ with $|z-z_0|<\delta$. The <triangle inequality> gives $|f(z)-f(z_0)|<\epsilon$. This proves the <uniform limit theorem> on the relative topology of $E$, whether or not $E$ is open.
The <Weierstrass M-test> states that if $|u_n(z)|\le M_n$ for every $z\in E$ and $\sum M_n<\infty$, then $\sum u_n$ converges uniformly and absolutely on $E$. Indeed, the supremum of a tail is at most the corresponding tail of $\sum M_n$, proving the uniform Cauchy criterion.
For the given <secant function> terms, a denominator can vanish only at $z=n(2k+1)$, an integer. Thus every individual term is continuous off the integers. Fix $z_0\notin\mathbb Z$, and choose a closed <disk> $K$ about $z_0$ containing no integer. Let $R=\sup_K|\operatorname{Re}z|$ and take $N>2R$. For $n\ge N$, $|\pi\operatorname{Re}z/(2n)|<\pi/4$; the permitted cosine inequality gives
$$
\left|n^{-2}\sec\frac{\pi z}{2n}\right|\le\frac{\sqrt2}{n^2}\quad(z\in K).
$$
The <Weierstrass M-test> makes the tail uniformly convergent on $K$. The finitely many earlier terms are continuous there, so the <uniform limit theorem> makes their sum continuous at $z_0$. Therefore \b[$f$ is continuous on $\mathbb C\setminus\mathbb Z$]. The argument is local; a global M-test on the whole punctured plane would fail near its <poles>.
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