Is my proof below valid? Thank you.
$\def\n{{\mathbf n}} \def\x{{\mathbf x}} \def\w{{\mathbf w}} \def\r{{\mathbf r}} \def\y{{\mathbf y}} \def\a{{\mathbf a}} \def\b{{\mathbf b}} \def\p{{\mathbf p}} \def\X{{\mathcal X}} \def\Y{{\mathcal Y}} \def\T{{\mathcal T}} \def\S{{\mathcal S}} \def\O{{\mathcal O}} \def\R{{\mathbb R}} \def\Rhat{{\widehat{\R}}}$
Let $\|\cdot\|_a$ and$\|\cdot\|_b$ represent two norms on $\R^n$ and suppose the norms are equivalent, that is, there exist constants $0<c_1 \le c_2 <\infty$ such that for all $\x\in\R^n$, $c_1\|\x\|_a \le \|\x\|_b \le c_2\|\x\|_a$. Prove that every set that is open with respect to $\|\cdot\|_a$ is also open with respect to $\|\cdot\|_b$. (Note: The set $\O\subseteq\R^n$ is open with respect to $\|\cdot\|_a$ if for every $\x\in\O$, there exists $\epsilon>0$ such that the set $\{\y\colon\|\x-\y\|_a<\epsilon\}\subseteq\O$. Similarly, $\O$ is open with respect to $\|\cdot\|_b$ if for every $\x\in\O$, there exists $\epsilon>0$ such that the set $\{\y\colon\|\x-\y\|_b<\epsilon\}\subseteq\O$.)
$\textbf{Solution:}$ editing
I believe that after the discussion in the comments, the proof is clear and complete. As a last remark, in the third line of the solution I'd probably prefer "$...||x-y||_b \le c_2||x-y||_a$ ,and/hence if$...$".