Suppose $N$ is a countably infinite index set of unique integers and $f_n\to f$ uniformly on some domain $D$ with $f_n,f$ holomorphic and $n\in N$. Suppose $N_1\subset N$ is a second countably infinite index set of unique integers. Then does $~f_n\to f$ uniformly where $n\in N_1$?
I think it should be true, since in my original proof of the uniform convergence of $f_n\to f$ with $n\in N$ I have shown that $|f-f_n|<\varepsilon$ for all $\varepsilon>0$ and $n>M$ for some $M>1$. If I use $n\in N_1$ instead then since $n\in N_1\subset N$, I kind of see that this should still work.
I suppose that, when you wrote than $N$ “is a countably infinite index set of unique integers”, what you meant was that $N$ is an infinite set of non-negative integers. Then, if $N_1\subset N$ and $N_1$ is infinite too, then the $k^\text{th}$ element of $N_1$ is greater than or equal to the $k^\text{th}$ element of $N$. Therefore, given $\varepsilon>0$, since you know that $|f-f_n|<\varepsilon$ if $n\geqslant M$ for a certain integer $M$ and $n\in N$, then the same thing is true if $n\in N_1$ and $n\geqslant M$.
So, the answer is affirmative.