Proof of Lemma 13.1 in Munkres

Question

I don't understand the last step of the following proof. Could someone please explain it to me?

Lemma 13.1: Let $X$ be a set; let $\mathcal{B}$ be a basis for a topology $\mathcal{T}$ on $X$. Then $\mathcal{T}$ equals the collection of all unions of elements of $\mathcal{B}$.

Proof: The elements of $\mathcal{B}$ are elements of $\mathcal{T}$. Because $\mathcal{T}$ is a topology, the union of elements of $\mathcal{B}$ are also elements of $\mathcal{T}$.

Conversely, given $U\in\mathcal{T}$, choose for each $x\in U$ an element $B_x$ of $\mathcal{B}$ such that $x\in B_x\subset U$.

Then $U=\bigcup_{x\in U}B_x$, ----------------------------------------------- [How??]

so $U$ equals a union of elements of $\mathcal{B}$.

Answer 1

You have $U \subseteq \bigcup_{x\in U}B_x$ as for $x \in U$, it exists $B_x \in \mathcal B$ such that $x\in B_x\subseteq U$.

Conversely all the $B_x$ are included in $U$. Therefore their union is too.