 Solution

Fix $z_0\in E$ and $ϵ>0$. Choose $N$ so that ${\sup }_E|f_N-f|<ϵ/3$, and then choose $\delta >0$ so that $|f_N(z)-f_N(z_0)|<ϵ/3$ for $z\in E$ with $|z-z_0|<\delta$. The triangle inequality gives $|f(z)-f(z_0)|<ϵ$. 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 $R={\sup }_K|\mathrm{Re}z|$ and take $N>2R$. For $n\ge N$, $|\pi \mathrm{Re}z/(2n)|<\pi /4$; the permitted cosine inequality givesThe 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 $f$ 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.