I think it will help to understand exactly what your after when considering a subbase. Start with choosing a random collection of subsets $S$ of $X$. In general, you know that this collection $S$ does not form a topology. Now comes the fundamental question:
"What do we have to change about $S$ to get a topology?"
Well, for one, arbitrary unions and finite intersections of elements of $S$ ought to again be in $S$. If they are not, we simply add them to $S$ to get a new collection $T$.
Now we also need $X$ and the empty set $\emptyset$ to be in $T$ as well. However, if we add the condition that the union of elements of $S$ is $X$, we know that $X$ will be in $T$ by construction. This is what motivates the definition of a subbase.
In your example, $T=\{X,\emptyset, \{c,d \}, \{b\}\}$ is not a topology since the union
$$\{c,d\}\cup \{b\}=\{c,b,d\}\not \in T.$$
Instead, let's consider
$$T=\{\emptyset, \{b\}, \{c,d \}, \{c,b,d\}, X\}.$$
Does this form a topology? Check the axioms. Is your original example
$$S=\{X,\emptyset, \{c,d \}, \{b\}\}$$
a subbase for T?
Here is your confusion: what Munkres starts out with is a set $X$ and a collection $\mathfrak B$ of subsets of $X$, which satisfy the given properties. Only then do we define the topology $\mathscr T$ using this basis, so until this is done it doesn't make sense to talk about the basis elements of $\mathfrak B$ as being "open".
Now, once we do define our topology, recall that we define a subset $U$ of $X$ to be open if given $x\in U$ there exists $B\in\frak B$ such that $x\in B\subseteq U$. Since this certainly holds when $U=B\in\frak B$, we see that all elements of $\frak B$ are open.
This is the precise statement of the lemma you are referring to:
Let $X$ be a set; let $\frak B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\frak B$.
So when Munkres says this, he is doing as we say above: that is, he is starting with some collection of subsets $\frak B$ that generate a topology, and he is calling this topology $\scr T$. So by what we said above, all basis elements are open in $\scr T$ and there should be no issue.
Best Answer
In this picture, the author was trying to prove that if you choose any two elements from a Topology $\tau$, then the intersection of that two sets is also in $\tau$. He basically proved that we can define topology by the definition of basis.
Also, note that the graphs in that picture are some regions, not the curve. The lemma tells us: If we choose any open set $O \in\tau$, then $O$ is the union of some basis elements. For example, any open set (does not matter how bad you can imagine) in $R^2$ with the standard topology is the union of some open discs.
Finally, please have a look at the example $1$ and $2$ of Munkress page $78$ (2nd edition). Here he showed two different bases $\mathbb{B}$ and $\mathbb{B^1}$ of the Euclidean topology of $\mathbb{R^2}$ as the collection of circular regions(interiors of the circles) and Open rectangles.
Answer to the last question:
Yes, any element of the basis $\mathbb{B}$ is open. More importantly, the rest of the open sets can be created by the union of the basis elements of that given basis $\mathbb{B}$.