$ ($int $S)\, \cup \,($int $T) \subseteq $ int$(S \cup T) $ 
For $S,T \subseteq \mathbb{R}^n$ prove that $ ($int $S)\, \cup \,($int $T) \subseteq $ int$(S \cup T) $ 

$($ int $S)$ represents the set of all interior point of $S$. The questions seems to be straight forward but I am not very confident in my proof. Here is what I did: 
Consider any $x \in ($int $S)\, \cup \,($int $T) $, which means $x \in ($int $S) $ or $x \in ($int $T)$ . This implies $ \exists \,B_{\delta_1}(x)$ and $B_{\delta_2}(x)$ for $\delta_1 , \delta_2 > 0$ such that $B_{\delta_1}(x) \subseteq S$ or $B_{\delta_2}(x)\subseteq T$. Choose $\delta $ = Min$(\delta_1,\delta_2)$. Then  $B_{\delta}(x) \subseteq S$ or $B_{\delta}(x)\subseteq T \implies B_{\delta}(x) \subseteq S \cup T $. Hence $x \in $ int $(S\cup T)$. 
Is this proof correct? Is there any other better way to solve this problem ?  
 A: $Int(S)$ and $Int(T)$ are open sets. Thus $Int(S) \cup Int(T)$ is an open set, contained in $S \cup T$. Since $Int(S \cup T)$ is the largest open set contained in $S \cup T$, $Int(S) \cup Int(T) \subseteq Int(S \cup T)$. 
A: Looks correct to me.
You don't need two $B$'s, though. There is a $B_\delta(x)$ such that either $B_\delta(x)\subseteq S$ or $B_\delta(x)\subseteq T$. Thus $B_\delta(x)\subseteq S\cup T$.
A: There is a solution for general proof of your question in this site: HERE

A counter example why the equality does not hold:
Let  $H_1 = \left[{0 , \dfrac 1 2}\right]$ and $H_2 = \left[{\dfrac 1 2 , 1}\right]$
$\displaystyle \left({H_1 \cup H_2}\right)^\circ$=$\displaystyle \left({\left[{0 , \dfrac 1 2}\right] \cup \left[{\dfrac 1 2 , 1}\right]}\right)^\circ$=$\displaystyle \left[{0 , 1}\right]^\circ$=$\displaystyle \left({0 , 1}\right)$
and
$\displaystyle {H_1}^\circ \cup {H_2}^\circ$=$\displaystyle \left[{0 ,, \dfrac 1 2}\right]^\circ \cup \left[{\dfrac 1 2 \, 1}\right]^\circ$=$\displaystyle \left({0 , \dfrac 1 2}\right) \cup \left({\dfrac 1 2 , 1}\right)$$\neq \displaystyle \left({0 , 1}\right)$
A: For any $S$,define $O(S)$ as the set of all open subsets of $S$.
Obviously $O(S)$ and $O(T)$ are subsets of $O(S\cup T)$ so $O(S)\cup O(T)\subseteq O(S\cup T).$
And $int (S)=\bigcup O(S).$
We have $$int (S)\cup int (T)=(\,\bigcup O(S)\,)\cup (\,\bigcup O(T)\,)=$$ $$=\bigcup (O(S)\cup O(T))\subseteq$$ $$\subseteq \bigcup O(S\cup T)=$$ $$=int (S\cup T).$$ 
A set $S$ is open iff $S=int(S).$ If $S$ and $T$ are open then $S\cup T$ is also open and $int(S)\cup int(T)=S\cup T=int (S\cup T).$
A common textbook example where $int(S)\cup int (T)\ne int (S\cup T)$ is, in the space $\Bbb R$ with the standard topology, $S=\Bbb Q$ and $T=\Bbb R\setminus \Bbb Q.$ So $int (S)$ and $int (T)$ are empty but $int (S\cup T)=int (\Bbb R)=\Bbb R.$
If you are not familiar with \bigcup ($\bigcup$): When $F$ is a set of sets, we define $\bigcup F=\cup_{f\in F}\,f$. That is, $x\in \bigcup F$ iff $x$ belongs to at least one member of $F$. So if $F\subseteq G$ then $\bigcup F\subseteq \bigcup G.$ In particular let $F=O(S)\cup O(T)$ and $G=O(S\cup T).$
