Topology and its basis In Munkres, Lemma 13.1 says that topology is the collection of all unions of elements of collection of basis elements.
Now in the figure 13.3, which is an irregular curve and also an open set i.e., element of topology, how to obtain that by union of basis elements that is circles?
I just learnt about that lemma and tried to visualise it using the figure(I know why that figure is there.)
 A: 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 $\mathbb{B}$.
A: The figure 13.3 is to illustrate the proof that $U_1,U_2 \in \mathcal{T}$, $U_1 \cap U_2 \in \mathcal{T}$. Here $U \in \mathcal{T}$ iff for each $x \in U$ we have a $B \in \mathcal{B}$ such that $x \in B \subseteq U$.
Maybe you're confused about the equivalence between "$U \in \mathcal{T}$ iff for all $x \in U$ we have a $B \in \mathcal{B}$ such that $x \in B \subseteq U$" and "$U\in \mathcal{T}$ iff $U$ is a union of elements from $\mathcal{B}$"? These are the two equivalent ways to define the topology $\mathcal{T}$ from a base $\mathcal{B}$.
See also this recent thread for some more discussion. The fact that both open squares as open discs form a base for the Euclidean plane, does indeed mean that open discs are (countable!) unions of open squares and vice versa.
