Collection of open sets is a basis for topology 
Let $X$ be a topological space. and let $\mathscr C$ a collection of open sets such that for each open set $U\subset X$ and for all $x\in U$ there is a open set $C\in \mathscr C$ such that $x\in C \subset U$. show that $\mathscr C$ is a basis for the topology on $X$

I tried to use the definition of basis:

Definition 2.1. Let $X$ be a set. A collection of set $\mathcal B\subseteq\mathcal P(X)$ is called a basis on $X$ if the following two properties are satisfied:
  
  
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*$\mathcal B$ covers $X$. That means: $\forall x in X$, $\exists B\in\mathcal B$ such that $b\in B$.
  
*$\forall B_1,B_2\in\mathcal B$, $\forall x\in B_1\cap B_2$, $\exists B\in\mathcal B$ such that $x\in B\subseteq B_1\cap B_2$.
  
  
  In words, the second property says: given a point $x$ in the intersection of two elements of basis, there is some element of the basis containing $x$ and contained in this intersection.


But I can't see why $\mathscr C$ covers $X$.
Any hints how can I show that?
 A: Take an arbitrary $x \in X$. Then $x \in X \subseteq X$ (using $X$ itself as an open subset of $X$). By definition of $\mathscr C$ there is a $C \in \mathscr C$ with $x \in C$. Hence $\mathscr C$ covers $X$.
A: There are two related notions here: 
We can start with a set $X$, and define a collection $\mathscr{B}$ satisfying the two properties, then we can define a smallest topology $\mathscr{T}$ which is the minimal topology that contains all members of $\mathscr{B}$. One possible definition is $\mathscr{T} = \{\cup \mathscr{B}': \mathscr{B}' \subseteq \mathscr{B}\}$, all unions of subfamilies of $\mathscr{B}$. One can show that this collection $\mathscr{T}$ is a topology iff $\mathscr{B}$ obeys the two conditions ($X \in \mathscr{T}$ needs the first, and $U,V \in \mathscr{T} \rightarrow U \cap V \in \mathscr{T}$ needs the second).  I could call this an "a priori" base.
On the other hand we can start with a topologial space $(X, \mathscr{T})$ and define that $\mathscr{B}$ is a base for the (pre given) topology $\mathscr{T}$ by demanding that for $\mathscr{B} \subset \mathscr{T}$ (all base members are open) and for all $O$ open, and all $x \in O$, there exists some $B \in \mathscr{B}$ such that $x \in B \subseteq O$. This condition is equivalent in fact to the fact that every open $O$ is a union of a subfamily of $\mathscr{B}$. The fact that this $\mathscr{B}$ also obeys the two conditions is clear as $X$ being open implies the first and the intersection of two base elements (which are themselves open) being open implies the second. I could call this base an " a posteriori" base
So in fact the $\mathscr{B}$ from the first type is a indeed a base for the topology $\mathscr{T}$ in the sense of the second type, justifying its name.
A: Let $U$ be an open set of $X$.
For each $x \in U$, by definition of $\mathscr{C}$, there is $C_x \in \mathscr{C}$ such that $x \in C_x \subseteq U$.
So $U = \bigcup_{x \in U} C_x$ is an union of elements from $\mathscr{C}$, and therefore, $\mathscr{C}$ is a basis for the topology of $X$.
