= Solution
Fix $x_0\in E$ and $\varepsilon>0$. By <uniform convergence>, choose $N$ so that $|f_N(x)-f(x)|<\varepsilon/3$ 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 $|f_N(x)-f_N(x_0)|<\varepsilon/3$. The <triangle inequality> gives
$$
|f(x)-f(x_0)|
\le |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|
<\varepsilon.
$$
This 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> and
$$
0<n^{-1-x}\le n^{-1-a},\qquad
\sum_{n=1}^\infty n^{-1-a}<\infty.
$$
The <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, \b[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 $x\downarrow0$ is bounded below by arbitrarily long tails of the divergent harmonic series. There is no continuity claim at the excluded endpoint $0$.
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