Prove that there is a unique topology is the coarsest containing topologies $T_\alpha$ $\{T_\alpha\}$ is a family of topologies on $X$. Prove that there is a unique
coarsest topology containing topologies $T_\alpha$ and a unique finest topology contained in all topologies $T_\alpha$.
 A: On the second part:
Let $\tau$ denote a topology such that it is contained in every $T_{\alpha}$. 
So denoting the index set by $A$ we have:$$\tau\subseteq\bigcap_{\alpha\in A} T_{\alpha}\tag1$$
Fortunately it can be shown that $\bigcap_{\alpha\in A} T_{\alpha}$ is a topology itself (try to prove that yourself) and $(1)$ tells us that any topology that is contained in every $T_{\alpha}$ is coarser (i.e. has no more elements). 
This allows us to conclude that $\bigcap_{\alpha\in A} T_{\alpha}$ is the finest topology that is contained in every $T_{\alpha}$.

On the first part.
As said in the comments the intersection of a collection of topologies on $X$ is a topology itself. 
If $\mathcal V\subseteq\wp(X)$ then we can take the collection of all topologies that contain $\mathcal V\subseteq\wp(X)$ and then take the intersection of these topologies.  Denoting this intersection by $\tau_{\mathcal V}$ we observe that:


*

*$\tau_{\mathcal V}$ is a topology.

*$\mathcal V\subseteq\tau_{\mathcal V}$

*If $\rho$ is a topology with $\mathcal V\subseteq\rho$ then  $\tau_{\mathcal V}\subseteq\rho$


This together tells us that $\tau_{\mathcal V}$ is the coarsest topology that contains $\mathcal V$.
You are searching for the coarsest topology that contains every $T_{\alpha}$ or equivalently for the coarsest topology that contains $\bigcup_{\alpha\in A}T_{\alpha}$, so you can apply this.
