A characterization of open sets using nets Dudley: Real Analysis and Probability - P31
2.1.3. Theorem  Let $(S,\mathcal T)$ be any topological space. Then
(b) A set $F\subset S$ is closed iff for every net $x_i\to x$ in $S$ with $x_i\in F$ for all $i$ we have $x\in F$.
(c) A set $U\subset S$ is open iff for every $x\in U$ and net $x_i\to x$ there is some $j$ with $x_i\in U$ for all $i\ge j$.
Proof.
(c) "If": suppose a set $B$ is not open. Then for some $x\in B$, by (b) there is a net $x_i\to x$ with $x_i\notin B$ for all $i$.
It seems easy but I just don't understand the last sentence in the proof. How can we find this kind of $x$ in $B$? I really appreciate any help! 
 A: A set $B
\subseteq$ is open if and only if its complement $B^c=S-B$ is closed. So if $B$ is not open, then $B^c$ is not closed.
Next by (b), since $B^c$ is not closed, it is not the case that "if $(x_i)$ is a net such that $x_i \to x$ with $x_i \in B^c$, then $x \in B^c$". Since that statement is false, there exists a convergent net $x_i$ in $B^c$ such that $x^i \to x$ but $x\in B$.
And this is the lsat sentence in your proof. The existence of such an $x$ follows from negating the statement in (b).
A: If B is not open, it is not true that every point has a neighborhood $V_x$ such that $V_x \subset B$.
Thus the $x \in B$ being referred to is one such x (i.e. x has no neighborhood that is contained in B; such a point exists if and only if B is open, by definition of openness).
Then as for the net, just choose the standard direction on the set of all neighborhoods of x, and then for each i it will be possible to choose a point in the ith neighborhood which is NOT in B (since none of the neighborhoods are properly contained in B). Therefore a sequence as claimed ($x_i \rightarrow x$, $x_i \not\in B$ for all i) exists.
A: Noting $F$ the complement of $U$ in $S$, you can see that (by definition):
 '$B$ is not open' implies '$F$ is not closed' which gives the last sentence
