Denote by $I$ the interval $[0,1]$. The space $\{0,1\}^I$ can be viewed as the space of all functions $$f: I \to \{0,1\}.$$ With the product topology (or equivalently, the point to point convergence topology), $\{0,1\}^I$ is known to be compact but not sequentially compact.
An example of a sequence without convergent subsequence is $f_n$, where $f_n(x)$ is the n-th binary digit of $x$ (where the number of ones cannot be finite, to avoid different binary representations).
Now, there is no subsequence $n_k$ such that $f_{n_k}(x)$ converges for all $x$. Just take an $x$ such that its binary expansion has the $n_k$-th digit equals to $0$ when $k$ is even and $1$ otherwise.
But since $\{0,1\}^I$ is compact, $f_n$ is supposed to have a convergent subnet! I would like to see such a subnet. :-)
Can you help me with that?
Pick an ultrafilter $F\subset \mathcal P(\mathbb N).$ Let $f$ be the function that sends $r$ to $1$ if and only if $\{n\mid f_n(x)=1\}\in F.$ Note $F$ is a directed set under inclusion.
I will argue that the subnet defined by $f'_{S}=f_{\min(S)}$ for $S\in F$ converges to $f.$ Any basic open set containing $f$ is of the form $U=\{g\mid g(r_i)=f'(r_i)\text{ for all $1\leq i\leq k$}\}.$ The sets $\{n\mid f_n(r_i)=f'(r_i)\}$ are in $F,$ so their intersection $S$ is in $F,$ and any $n\in S$ satisfies $f_n\in U.$ So $f'$ is eventually in $U,$ which is the definition of a limit.