Criterion for a basis of a topology Here is part of Munkres' book on topology:
I suppose the reader is supposed to use

to prove this. But I don't understand how. The lemma says that if there is a space with a fixed topology, then to show that a set forms a basis, one has to take an open set from the fixed topology and do something. But the order topology is being defined, how can I take an open set and prove that it contains a basis element if I don't know what open sets are yet?
 A: $\mathscr B$ doesn't define a topology on $X$.  You need to take arbitrary unions and finite intersections of elements of $\mathscr B$ to fulfill the definition of a topology.  
$\mathscr B$ is a basis for the topology it generates...
Actually,  $\mathscr B$ is closed under finite intersections, as Munkres notes.
But you need to throw in arbitrary unions...  $\mathscr B$ isn't closed under arbitrary unions. 
Thus $\mathscr B$ generates a topology on $X$.
A: The requirements for $\mathcal{B}\subseteq \mathscr{P}(X)$ to be a base for some topology on $X$ are (second edition., paragraph 13, page 78):
(1) $\bigcup \mathcal{B} = X$, or equivalently for every $x \in X$ there is some $B \in \mathcal{B}$ such that $x \in B$.
(2) For all $B_1, B_2 \in \mathcal{B}$, for all $x \in B_1 \cap B_2$: there exists some $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$.
The lemma 13.2 is not used here: this is to characterise a base for an already given topology, while in the case at hand we want to define a topology by a suitable base.
Munkres suggests to check condition (2) by checking that the set of intervals he describes ($\mathcal{B}$) has the property that if $B_1 , B_2 \in \mathcal{B}$ we either have $B_1 \cap B_2 \in \mathcal{B}$ again (so we can take $B_3 = B_1 \cap B_2$) or $\emptyset$ (in which case the condition is voidly true as there are no $x \in B_1 \cap B_2$ to check.
