An Explicit Example of a Sequence Which has Convergent Subnet but no Convergent Subsequence

I get confused by my (faked) intuition that all subnets of a sequence should also be a subsequence. This turns out to be wrong, and I try to seek for easy example.

The following is the easiest one:

Example. Let $\mathcal I$ be the set of vectors $(n_1,n_2,\dots)$, where $n_i\in \N$ for each $i$ and $n_1<n_2<\cdots$. The cardinality of $\mathcal I$ is easily seen to be $|\R|$. Now our desired sequence will be a sequence of functions defined on $\mathcal I$ as follows:

Let $i=(n_1,n_2,\dots)\in \mathcal I$, we define \[
f_n(i)=\begin{cases}
(-1)^k,&\text{if }n=n_k,\\
0,&\text{otherwise}.
\end{cases}
\] It is easy to see that given an $i=(n_1,n_2,\dots)\in \mathcal I$, $\{f_{n_k}(i)\}=\{(-1)^k\}$ diverges.

With a little abuse of notation, we define $f_n = (f_n(i))_{\mathcal I}$. Now $f_1,f_2,\dots \in [-1,1]^{\mathcal I}$. By Tychonoff's Theorem $f_1,f_2,\dots \in [-1,1]^{\mathcal I}$ is compact w.r.t. the product topology, therefore $\{f_n\}$, being a net, must have a convergent subnet by a standard exercise on nets.

We claim that $\{f_n\}$ has no convergent subsequence. Suppose it does, then there is $\{n_k\}$ such that $\{f_{n_k}\}$ converges w.r.t. product topology. By definition, it has coordinatewise convergence: For every $i\in \mathcal I$, $f_{n_k}(i)$ converges. This is a contradiction if we choose $i=(n_1,n_2,\dots)$.$\qed$