Exercise 10(a), Section 19 of Munkres’ Topology 
Let $A$ be a set; let $\{X_\alpha \}_{\alpha \in J}$ be an indexed family of spaces; and let $\{ f_\alpha \}_{\alpha \in J}$ be an indexed family of functions $f_\alpha \colon A \to X_\alpha$.
(a) Show there is a unique coarsest topology $\mathscr{T}$ on $A$ relative to which each of the functions $f_\alpha$ is continuous.

I can’t solve this problem. Maybe I don’t understand “exactly” what to prove. For instance, closure $\overline{A}$ is the smallest closed subset of $X$ containing $A$ means (1) $\overline{A}$ is closed in $X$ and it contains $A$, and (2) if $A\subseteq V$ where $V$ is closed in $X$ then $\overline{A} \subseteq V$. You can also give hint(s) to solve this problem.
Edit: The following definition I sto.. ahaa aaa took from Henno Brandsma answer

Let $(X,\mathcal{T})$ be a topological space.
Let $I$ be an index set, and let $Y_i (i \in I)$ be topological spaces
and let $f_i: X \rightarrow Y_i$ be a family of functions.
Then $\mathcal{T}$ is called the initial topology with respect to the maps $f_i$
iff

*

*$\mathcal{T}$ makes all $f_i$ continuous.

*If $\mathcal{T}'$ is any other topology on $X$ that makes all $f_i$ continuous, then $\mathcal{T} \subseteq \mathcal{T}'$.



This is the so-called initial topology on $A$ wrt the maps $f_\alpha, \alpha \in J$. See my post here for an existence proof and all details and applications of this notion.
Short answer: it's the topology on $A$ with subbase $$\{f_\alpha^{-1}[O]\mid j \in J ; O \subseteq X_\alpha \text{ open }\}$$
