 Solution

Fix $x_0\in E$ and $\epsilon >0$. By uniform convergence, choose $N$ so that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in E$. By continuity of $f_N$ relative to $E$, there is $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality givesThis proves the uniform limit theorem without requiring $E$ to be open or closed.

For the series, fix an arbitrary $x_0\in (0,1]$ and put $a=x_0/2>0$. On $[a,1]$ each summand $n^{-1-x}$ is continuous andThe Weierstrass M-test gives uniform convergence of the series on $[a,1]$. Its partial sums are continuous, so the uniform limit theorem makes its sum continuous on that interval, in particular at $x_0$ with the appropriate relative topology if $x_0=1$. Since $x_0$ was arbitrary, the sum is continuous on $(0,1]$.

This is local uniform convergence, not uniform convergence on the entire half-open interval. Indeed, for any finite partial sum length $N$, its tail as $x\downarrow 0$ is bounded below by arbitrarily long tails of the divergent harmonic series. There is no continuity claim at the excluded endpoint $0$.