Subnets vs. Subsequences

Take $X = \{0,1\}^{[0,1]}$, thought of as the set of all functions from the unit interval $[0,1]$ to the two-point set $\{0,1\}$. Equip $X$ with the product topology; $X$ is compact by Tychonoff's theorem. In the product topology, a sequence $f_n$ converges iff it converges pointwise, i.e. if the $\{0,1\}$-valued sequence $f_n(x)$ converges (i.e. is eventually constant) for every $x \in [0,1]$.

Define $f_n : [0,1] \to \{0,1\}$ to be the function such that $f_n(x)$ is the $n$th bit in the binary expansion of $x$. (If $x$ has more than one binary expansion, take $f_n(x) = 0$ for all $n$; it doesn't really matter here.) Then given any subsequence $f_{n_m}$, you should be able to produce an $x \in [0,1]$ such that $f_{n_m}(x)$ does not converge. Hence $\{f_n\}$ has no convergent subsequence.

My favorite example is $\beta\omega$, in which the sequence $\langle n:n\in\omega\rangle$ has no convergent subsequence. To see this, let $A$ be any infinite subset of $\omega$, and suppose that $\langle n:n\in A\rangle\to p$ for some $p\in\beta\omega$. Clearly $p\in\beta\omega\setminus\omega$, so $p$ is a free ultrafilter on $\omega$. Partition $A$ into two infinite subsets $A_0$ and $A_1$. Since $p$ is an ultrafilter, exactly one of $A_0$ and $A_1$ belongs to $p$, say $A_0$. But then $A_0\cup\{q\in\beta\omega\setminus\omega:A_0\in q\}$ is an open nbhd of $p$ that misses infinitely terms of $\langle n:n\in A\rangle$, so $\langle n:n\in A\rangle$ that does not converge to $p$ after all.