
clousre
Suppose that $\displaystyle \displaystyle \tau 1 \mbox and \quad \tau2 \mbox \quad are \quad \quad \quad topologies \quad on \quad a set X \quad with \quad \tau1 \subset \tau2 . \mbox \quad if A \subset X \quad$
$\displaystyle \mbox prove \quad that : \quad \overline{A}^2 \subset \overline {A}^1 \mbox \quad where \overline {A}^1 \mbox is \quad the \quad clousre $ $\displaystyle \quad of A \quad in (X, \tau1 ) \quad \overline {A}^2 \mbox is \quad the \quad clousre \quad of A \quad in (X,\tau2) $

If $\displaystyle A$ is an open subset in $\displaystyle \tau_1$ then $\displaystyle A$ is open in $\displaystyle \tau_2$. So, if $\displaystyle B$ is closed in $\displaystyle \tau_1$ then $\displaystyle B$ is closed in $\displaystyle \tau_2$. Now you define the closure of $\displaystyle A$ in $\displaystyle \tau_i$ as the intersection of all closed subsets of $\displaystyle (X, \tau_i)$ such that they contain $\displaystyle A$, but since closed sets in $\displaystyle \tau_1$ are closed in $\displaystyle \tau_2$ then the closure in $\displaystyle \tau_2$ has more elements to intersect and thus is contained in the closure in $\displaystyle \tau_1$.
Well, that is an overview of the proof. I leave the details to you.

Here is a different approach.
$\displaystyle x \in \overline A $ if and only if each for each open set $\displaystyle O$, $\displaystyle x \in O\, \Rightarrow \,O \cap A \ne \emptyset $.
But $\displaystyle O \in \tau _1 \; \Rightarrow \;O \in \tau _2 $.