How are basis elements also elements of the topology? I read the following definitions in Munkres' Topology (2nd Edition):

If $X$ is a set, a basis for a topology on $X$ is a collection
  $\mathcal B$ of subsets of $X$ (called basis elements) such that
  
  
*
  
*For each $x\in X$, there is at least one basis element $B$ containing $x$.
  
*If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that
  $B_3\subset B_1\cap B_2$.
  
  
  If $\mathcal B$ satisfies these two conditions, then we define the
  topology $\mathcal T$ generated by $\mathcal B$ as follows: A subset
  $U$ of $X$ is said to be open in $X$ (that is, to be an element of
  $\mathcal T$) if for each $x\in U$, there is a basis element
  $B\in\mathcal B$ such that $x\in B$ and $B\subset U$. Note that each
  basis element is itself an element of $\mathcal T$.

From this, I am having difficulties understanding how the last sentence follows. Could someone please point me in the right direction?

Just expanding on Daniel Fischer's comment a bit:
From the definition of an open set from a basis, we have:

A subset $U$ of $X$ is said to be open iff for each $x\in U$, there is a basis element $B\in\mathcal B$ such that $x\in B\subseteq U$.

So consider this:
Take a $B \in \mathcal{B}$ and further let $x\in B$.
There exists a basis element, namely $B$, such that:
$x \in B \subseteq B$,
and hence $B$ is open.
