This is infact the problem 4c in the book Topology of Munkres. This is my idea to prove it, could you please check for me whether it contains any mistake. I've already checked that there are various questions with the same title but different in purposes, so I still decided to post it here. Thank you.
Let $\{T_\alpha\}$ be a family of topologies on X. Show that there is a unique smallest topology on X containing all the collection $T_\alpha$.
Proof: Define the topology $\mathcal{T}$ generated by $\bigcup_{\alpha}\mathcal{T}_{\alpha}$ as following (the intersection is finitely taken):
$$\mathcal{T}= \left\lbrace \cap U_{i\alpha} \mid\ U_{i\alpha}\in \mathcal{T}_{\alpha} \right\rbrace $$Since each $T_\alpha$ is a topology then clearly $X; \emptyset\in \mathcal{T}$. If $U, V\in \mathcal{T}$ then $U=\cap U_{i\alpha}; V=\cap V_{j\alpha}$. So $U\cap V\in \mathcal{T}$ and
$$ U\cup V = \left(\cup U_{i\alpha}\right)\cap \left(\cup V_{j\alpha}\right) = \cup \left( U_{i\alpha}\cap V_{j\alpha}\right)$$
which also belongs to $\mathcal{T}$.
Thus $\mathcal{T}$ is a topology on $X$.
Let $\mathcal{T'}$ be any topology contains all $\mathcal{T}_{\alpha}$. Suppose $U$ is an open set in $\mathcal{T}$ then $U=\cap_{\alpha} U_{i\alpha}$. By the definition of $\mathcal{T'}$ then $U_{i\alpha}\in \mathcal{T'}$ hence $U\in \mathcal{T'}$. Therefore $\mathcal{T}\subset \mathcal{T'}$ so $\mathcal{T}$ is the smallest.
This is not correct. Think about the topologies $\tau_i$ on $X=2^{\mathbf N}$ where $\tau_i$ has exactly four open sets: $\emptyset$, $X$, $A_i=\{(x_n)\mid x_i=0\}$ and $A_i'=\{(x_n)\mid x_i=1\}$.
What is the finest topology containing each $\tau_i$, and what is the topology you try to describe?
There are other problems: you did not really say why your $\mathcal T$ it is closed under intersections, and you only showed that it is closed under binary, not arbitrary unions, and even that you did not do quite correctly, since you used unions instead of intersections (and vice versa).