Let $X$ be a set and let $\mathcal{B}$ be a basis for a topology on $X$. This means that $\mathcal{B}$ is a collection of subsets of $X$ such that for all $x \in X$ there is a basis element $B \in \mathcal{B}$ such that $x \in B$ and whenever $x$ is in the intersection of two basis elements $B_1$ and $B_2$, there is a third basis element $B_3$ such that $x \in B_3$ and $B_3 \subseteq B_1 \cap B_2$ (Munkres definition).
Here are what I understand to be two equivalent definitions of topologies generated by a basis. That is, one can generate a topology for $X$ from $\mathcal{B}$ as
\begin{equation} \mathcal{T}_{\mathcal{B}} = \{\bigcup C \mid C \subseteq \mathcal{B}\}. \end{equation}
Or, alternatively and equivalently, one can build the same topology by building the set
\begin{equation} \mathcal{T}_{\mathcal{B}}' = \{U \subseteq X \mid (\forall x \in U)(\exists B \in \mathcal{B})( x \in B \wedge B \subseteq U)\}. \end{equation}
Question 1) Is this the correct way to understand the material? That is, there is a definition of basis and two equivalent ways to build a topology from it?
Question 2) As far as I understand, the two sets should be equivalent. Online I have only found proofs of this equivalence by first proving that one of the two actually constitutes a topology (Munkres does this as well) and then showing the rest. Is this a necessary step? is it done out of convenience? is it sufficient to show that $\mathcal{T}_\mathcal{B} = \mathcal{T}_\mathcal{B}'$ as follows for instance?
Proof: Let $V \in \mathcal{T}_\mathcal{B}$. Then, $V$ is the union of some basis elements. Since all basis elements are subsets of $X$, their union is a subset of $X$; i.e,. $V \subseteq X$. Let now $x \in V$. Then, $x$ belongs to some basis element $B \in \mathcal{B}$ and that basis element must be fully contained in $V$. Thus, $V \in \mathcal{T}_\mathcal{B}'$. Let now $V \in \mathcal{T}_\mathcal{B}'$. I will show that this is the union of some basis elements. Consider some $x \in V$. By definition there is a $B_x \in \mathcal{B}$ such that $x \in B_x$ and $B_x \subseteq V$. Thus, $x \in \bigcup_{x\in V}B_x$. Suppose now $x \in \bigcup_{x \in V}B_x$. Then, $x \in B_x$ for some $x \in V$. Since all $B_x$ are subsets of $V$, then $x \in V$. Thus, $V = \bigcup_{x\in V}B_x$ and $V \in \mathcal{T}_\mathcal{B}$.
Everything is correct. In the first definition you need the convention that if $C = \emptyset$, then $\bigcup C = \emptyset$.
In your proof of $\mathcal{T}'_{\mathcal{B}} \subset \mathcal{T}_{\mathcal{B}}$ it would be better (see Anne Bauval's comment) to define $$C(V) = \{ B \in \mathcal B \mid B \subset V \} .$$ Then $V = \bigcup C(V)$. In fact, $\bigcup C(V) \subset V$ is trivial. To show $V \subset \bigcup C(V)$, consider $x \in X$. There exists $B_x \in \mathcal B$ such that $x \in B_x \subset V$. Thus $B_x \in C(V)$ and $x \in \bigcup C(V)$.