Let $A_1$ and $A_2$ Two sets on the metric space $(X, \rho)$ show that :
a) $Cl(A_1\cup A_2) = CL(A_1)\cup Cl(A_2)$
b) $A_1\subset A_2 \implies Cl(A_1)\subset Cl(A_2) $
This is my attempt: $$ \text{If } x\in Cl(A_1\cup A_2)\\ B_\epsilon(x)\cap (A_1\cup A_2)\neq\varnothing\\ (B_\epsilon\cap A_1)\cup (B_\epsilon \cap A_2)\neq \varnothing\\ B_\epsilon\cap A_1 \neq \varnothing \implies x\in Cl(A_1)\\ \text{or }B_\epsilon\cap A_2 \neq \varnothing \implies x\in Cl(A_2) $$ For the second part: $$ \text{let } x\in A_1\implies x\in A_2\\ \text{if }x\in Cl(A_1)\implies B_\epsilon(x)\cap A_1\neq \varnothing\\ \text{but }A_1\subset A_2\\ \implies B_\epsilon(x)\cap A_2\neq\varnothing\implies x\in Cl(A_2) $$