I have a open, bounded set $\Omega \subset \mathbb{R}^2$. For $h \geq 0$ one can create the following grid: $$\Omega_h = \{ x \in \Omega \mid x=(ih,jh), \quad i,j\in \mathbb{Z} \}.$$
It is claimed that for small enough $h$ the grid $\Omega_h$ is discrete connected, i.e. for each $x,y\in \Omega_h$ there is a path from x to y in $\Omega_h$, if $\Omega$ is connected.
I don't know how to show it, but I had the following Idea:
Consider the grid secquence $\Omega_{h_n}$ with $h_n = 1/2^n$. Then $\Omega_{h_n} \subset \Omega_{h_{n+1}}$. Assume in $\Omega_{h_n}$ are two disconnected points a,b(no path on grid from a to b).Then, if $\Omega$ is path connected, I find a continuous path from a to b. Using compactness I find a finite sequcene of $\varepsilon$ balls covering the path and lying in the grid. If I choose $h_m < \varepsilon/2$, then these two points are connected in $\Omega_{h_m}$. I'm actually not quite sure if this is true, but even if it true, there might be new points in the grid that are not connected. Either my idea does not work at all or I missed something. Can anyone help?
I think, I found a counterexample.
Define $$ t_n = \sum_{i=1}^{n-1} h_i. $$ Set $$ \Omega := \bigcup_{n=1}^\infty B_{\sigma h_n}(t_n,t_n) $$ with $\sigma \in (\frac23\sqrt2,\ 1)$.
Then $\Omega$ is open (union of open balls) and bounded. The mid-point of the $n$-th ball is in the discrete set $\Omega_{h_n}$. But there is no discrete connection to the $(n-1)$th ball.
To see that $\Omega$ is connected: the distance between the mid-point of the $n$-th and $(n+1)$th ball is equal to $$ \sqrt 2 h_{n}, $$ while the sum of their radii is $$ \sigma (h_n + h_{n+1}) = \sigma\left( h_{n} + \frac 12 h_{n} \right)= \sigma \frac32 h_n > \sqrt 2 h_n. $$