Let $\Omega$ be an open set in $\mathbb{C}$ and let $f$ be a (complex-valued) continuous function on $\Omega$. We also view $f$ as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ and write $f(x,y)=u(x,y)+iv(x,y)$. For each integer $k\geq 1$, define the sets
$\Omega_k^{(1)}=\left\{(x,y)\in\Omega|\text{ For all }(x',y)\in\Omega \text{ such that }|x'-x|\leq\frac{1}{k}, \text{ we have } |f(x',y)-f(x,y)|\leq k|x'-x| \right\}$
$\Omega_k^{(2)}=\left\{(x,y)\in\Omega|\text{ For all }(x,y')\in\Omega \text{ such that }|y'-y|\leq\frac{1}{k}, \text{ we have } |f(x,y')-f(x,y)|\leq k|y'-y| \right\}$
and let $\Omega_k=\Omega_k^{(1)}\cap\Omega_k^{(2)}$.
My question is: Why are the sets $\Omega_k$ closed (in $\Omega$)?
I thought that this follows from the continuity of $f$ and the fact that pre-image of a closed set under a continuous mapping is closed. So I tried writing $\Omega_k^{(1)}$ by $$\Omega_k^{(1)}=\bigcap_{\forall (x'y)\in\Omega}\left(\left\{(x,y)\in\Omega|\text{ Either } |x-x'|>\frac{1}{k} \text{ or } |f(x,y)-f(x',y)|\leq k|x-x'|\right\}\right)$$ but the problem was that the set in the parenthesis need not be closed!. (If it were, then $\Omega_k^{(1)}$ would be an intersection of a family of closed sets, hence closed.)
The book I'm reading just says that the sets $\Omega_k$ are closed, with no further explanations. Can someone show me why the sets are closed?
It's simpler to directly show that $\Omega \setminus \Omega_k^{(i)}$ is open.
If $(x,y) \in \Omega\setminus \Omega_k^{(1)}$, then by definition there is an $x'$ with $(x',y) \in \Omega$ and $\lvert x' - x\rvert \leqslant \frac{1}{k}$ and $\lvert f(x',y) - f(x,y)\rvert > k\lvert x' - x\rvert$. By continuity,
$$\lvert f(\xi + (x' - x), \eta) - f(\xi, \eta)\rvert - k\lvert x' - x\rvert$$
is positive on a neighbourhood $U$ of $(x,y)$, and that means $U \subset \Omega \setminus \Omega_k^{(1)}$. Thus $\Omega\setminus \Omega_k^{(1)}$ is open.
The argument for $\Omega_k^{(2)}$ is analogous.