Prob. 5, Sec. 13 in Munkres' TOPOLOGY, 2nd ed: How to distinguish between a basis and a subbasis? Here's Prob. 5, Sec. 13 in the book Topology by James R. Munkres, 2nd edition:

Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$. Prove the same if $\mathscr{A}$ is a subbasis.

Here is my solution:

First suppose that $\mathscr{A}$ is a basis for a topology $\mathscr{T}$ on a non-empty set $X$. The family of all the topologies on $X$ that contain $\mathscr{A}$ contains the discrete topology on $X$ is therefore non-empty; let $\mathscr{T}^\prime$ be the intersection of this family. We show that $\mathscr{T}$ equals $\mathscr{T}^\prime$.
Let $U$ be a set in $\mathscr{T}$. Then $U$ equals the union of some  subcollection of $\mathscr{A}$, and each set in that subcollection is in the topology $\mathscr{T}^\prime$, showing that $U$ is also in $\mathscr{T}^\prime$. Thus $\mathscr{T} \subset \mathscr{T}^\prime$. Am I right?
For the converse, we note that $\mathscr{A}$ is itself contained in the topology $\mathscr{T}$, and so $\mathscr{T}$ is in the family of all the topologies on $X$ that contain $\mathscr{A}$, and since $\mathscr{T}^\prime$ is the intersection of this family of topologies, therefore we can also conclude that $\mathscr{T}^\prime \subset \mathscr{T}$. Am I right?
Next we assume that $\mathscr{A}$ is a subbasis for topology $\mathscr{T}$. If $U$ is a set in $\mathscr{T}$, then $U$ is the union of some collection of sets each of which is the intersection of some finite subcollection of $\mathscr{A}$ and hence some finite subcollection of the topology  $\mathscr{T}^\prime$ since $\mathscr{T}^\prime$ contains $\mathscr{A}$, thus showing that $\mathscr{T} \subset \mathscr{T}^\prime$. Am I right?
The reverse inclusion follows by the same argument as above since $\mathscr{A}$ is still contained in $\mathscr{T}$. Am I right?

Now my question is, how do we distinguish (the necessity for) a subbasis for a topology from (the necessity for) a basis? Is there a subbasis that is not a basis, and vice versa?
We define a topology in terms of a basis because it is sometimes simply not possible to characterise the topology any other way. Am I right? If so, is there a similar reason why we talk of a subbasis?
Munkres simply hasn't touched upon these points, has he?
 A: There certainly are subbases that aren't bases. E.g. ordered spaces $(X,<)$ are best described using the subbase $\mathcal{S} = \{L_x, R_x: x \in X\}$, where $L_x = \{p \in X:  p < x\}$ and $R_x = \{p \in X: p > x\}$. (left segments and right segments).
On the reals (or most ordered sets) this is not a base as we cannot find an $L_x$ or $R_x$ sitting inside $(a,b)= L_b \cap R_a$, with $a < b$.
The corresponding base is the collection of all $L_x, R_x$ and open intervals (as intersections of 2 subbase elements), as we can reduce $L_a \cap L_b = L_{\min(a,b)}$ and similarly for any finite intersections of $L_x$'s. The same holds for finite intersections of $R_x$ using the maxima. So a finite intersection of sets of the form $L_x$ and $R_x$ reduce to at most 1 $L_x$ and at most one $R_y$, hence the intervals we get additionally.
Similarly for a product space $X = \prod_{a \in A} X_a$, the standard subbase is given by $\mathcal{S}= \{ p_\alpha^{-1}[U_\alpha]:  \alpha \in A, U_\alpha \subseteq X_\alpha \text{ open }\}$ and $p_\alpha: X \to X_\alpha$ are the projections. (this is also not a base, except for trivial cases, non-trivial finite intersections are open and don't contain a member from $\mathcal{S}$)
This allows us to say that 
1. The order topology is the smallest topology on $X$ that makes all left and right segments open.


*The product topology is the smallest topology that makes all projections continuous.


In both cases, we get relatively easy proofs of compactness in products and ordered spaces, using Alexander's subbase lemma:

Let $X$ be a space with a subbase $\mathcal{S}$. Then $X$ is compact iff every cover of $X$ by members of $\mathcal{S}$ has a finite subcover.

