${\bf exercise:}$ Let $X$ be a topological space. Suppose $\mathscr{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there is an element $C \in \mathscr{C}$ such that $x \in C \subset U$. Then, $\mathscr{C}$ is a basis for the topology of $X$.
Proof:
Let $x \in X$.Since $X$ is itself a topology of $X$, then $\exists C \in \mathscr{C}$ with $x \in C \subset X $. So, the first condition of basis is satisfied.
Next, suppose $x \in C_1 \cap C_2$ where $C_i \in \mathscr{C}$ We are given that this collection are open sets (elements of topology of $X$) so that $C_1 \cap C_2$ is open. Thus, there is $C_3 \in \mathscr{C}$ with $x \in C_3 \subset C_1 \cap C_2$
Therefore, $\mathscr{C}$ is a basis for the topology of $X$.
Now, I understand this is correct proof, but I am still confused as to why in the solution they include the following:
Let $\mathscr{T}$ be topology of $X$ and let $\mathscr{T}_2$ be topology generated by $\mathscr{C}$. We must show that $\mathscr{T} = \mathscr{T}_2$
Question: Why do we need to prove this? Isn't the work I showed more than enough to prove $\mathscr{C}$ is basis? I dont get it. Any help/clarificitation would be greatly appreciated.
$X$ is not a topology on $X$; I think that you meant that $X$ is an open set.
You have indeed shown that $\mathscr{C}$ is a base for some topology $\mathscr{T}_2$ on $X$. Moreover, if $\mathscr{T}$ is the topology on $X$, we know that $\mathscr{C}\subseteq\mathscr{T}$, so unions of members of $\mathscr{C}$ are in $\mathscr{T}$, and therefore $\mathscr{T}_2\subseteq\mathscr{T}$. But this does not guarantee that $\mathscr{T}_2=\mathscr{T}$, i.e., that $\mathscr{C}$ is a base for the original topology $\mathscr{T}$.
Let me give you an example. Let $X$ be $\Bbb R$ with the discrete topology $\mathscr{T}_d$, and let $\mathscr{C}$ be the set of open intervals $(a,b)$ with $a,b\in\Bbb R$ and $a<b$. Then it’s easy to show that $\mathscr{C}$ is a base for a topology on $\Bbb R$, using exactly the same argumentation that you used, but that topology isn’t the discrete topology: it’s the usual Euclidean topology $\mathscr{T}_E$. $\mathscr{T}_E\subseteq\mathscr{T}_d$ but the two aren’t equal: $\mathscr{T}_E\subsetneqq\mathscr{T}_d$.
To finish your argument, you need to show that $\mathscr{T}\subseteq\mathscr{T}_2$, so that you can conclude that in fact $\mathscr{T}_2=\mathscr{T}$. So let $U\in\mathscr{T}$. For each $x\in U$ there is a $C_x\in\mathscr{C}$ such that $x\in C_x\subseteq U$, and you can easily verify that $U=\bigcup_{x\in U}C_x\in\mathscr{T}_2$. Thus, every original open set is in the topology generated by $\mathscr{C}$, meaning that $\mathscr{T}\subseteq\mathscr{T}_2$, and we can now conclude that $\mathscr{T}_2=\mathscr{T}$, i.e., that $\mathscr{C}$ is not just a base for some topology on $X$, but in fact a base for the original topology $\mathscr{T}$ on $X$.