About $\mathring{A_1}\cup\mathring{A_2}\cup\mathring{A_3}=\mathring{\overbrace{A_1\cup A_2\cup A_3}}$

If $\operatorname{Bd}(A_1)\cap\operatorname{Bd}(A_2)\cap\operatorname{Bd}(A_3)=\emptyset$, show that $\operatorname{int}(A_1)\cup\operatorname{int}(A_2)\cup\operatorname{int}(A_3)=\operatorname{int}(A_1\cup A_2\cup A_3)$?

With $\operatorname{int}(A)$ is the interior if $A$, and $\operatorname{Bd}(A_1)$ is the boundary of $A_1$

Proof

Suppose $x\in\operatorname{int}(A_1\cup A_2\cup A_3)$ but $x\notin \operatorname{int}(A_1)\cup\operatorname{int}(A_2)\cup\operatorname{int}(A_3);$ I claim that $x\in\operatorname{Bd}(A_1)\cap\operatorname{Bd}(A_2)\cap\operatorname{Bd}(A_3).$ I will show that $x\in\operatorname{Bd}(A_1);$ the proof of $x\in\operatorname{Bd}(A_2)$ and $x\in\operatorname{Bd}(A_3)$ is similar. Since $x\notin\operatorname{int}(A_1),$ it will suffice to show that $x\in\operatorname{Cl}(A_1).$ Let $U$ be any neighborhood of $x.$ Then $V=U\cap\operatorname{int}(A_1\cup A_2\cup A_3)$ is a neighborhood of $x,$ and $V\subseteq A_1\cup A_2\cup A_3,$ but $V\not\subseteq A_2\cup A_3$ (since $x\notin\operatorname{int}(A_2))\cup\operatorname{int}(A_3)),$ so $V\cap A_1\ne\emptyset$ and therefore $U\cap A_1\ne\emptyset.$

is my Proof correct?

Any help will be appreciated.

No, counterexample in $\Bbb R$: $A_1 = [0,1], A_2= [1,2], A_3= [2,3]$.

Then $\bigcap_{i=1}^3 \text{Bd}(A_i) = \emptyset$. But the union of interiors does not contain $1$ and $2$ while the interior of their union does.

You'll probably be better off with $3$ conditions for the pairs of boundaries instead.