finite union of connected dense subsets of $\Bbb R^2$ For each $i\in\{1,2,\dots ,n\}$ define a set $$A_i:= \bigg\{(x,y)\colon x\in(k_i,k_i+1) , y\neq 0\bigg\}$$, to be dense and connected in $(k_i,k_i+1)\times\Bbb R$ where the $k_i$ are integer numbers . Put$$ M:=\bigg\{(x,0)\colon x\in\Bbb R\setminus\bigcup_{i=1}^{n}(k_i,k_i+1)\bigg\}.$$ Now, $$D:= \bigcup_{i=1}^{n} A_i \cup M$$ I am sure that $D$ is still a connected subset of $\Bbb R^2$ since the density of $A_i$ will ensure that they will have a limit point in common, By using the well known fact that says :If $B\subset\Bbb R^2$ is connected and $x$ is a limit of of $B$, then $B\cup \{x\}$ is still connected.
My question I need to see if my explanation is strong enough to show that $D$ is connected subset $\Bbb R^2$,  Thank you in advance.
Edit: The original was missing an important assumption for $A_i$ which is $y\neq 0$
 A: 1) this is the answer before you added the condition that $y \neq 0$ to the definition of the $A_i$:
Your statements about density and limit points aren't really the important things here. To see that $D$ is connected, show that it contains the $x$-axis, $X$ say, which meets each of the connected sets $A_i$. As the $x$-axis is connected this means that $D$ is connected, because you can write $D$ as:
$$
D = \bigcup_{i=1}^n (A_i \cup X)
$$
which displays $D$ as the union of a family of connected sets $X_i = A_i \cup M$ such that the intersections $X_i \cap X_j$ are all non-empty.
2) now you have added the condition that $y \neq 0$ to the definition of the $A_i$, the set $D$ remains connected:
To see this, let $D'$ be $D$ as it was without the restriction that $y\neq 0$ in the definition of the $A_i$. Then
$$
D = D' \setminus \bigcup_{i=1}^n (k_i, k_i + 1)\times \{0\}
$$
Removing these open line segments from $D'$ does not disconnect it, in the same way that removing a rung from a ladder (with at least two rungs) does not disconnect it).
To give a more formal proof, observe that you can replace each $A_i$ in the definition by:
$$
B_i:= \left\{(x,y)\colon x\in(k_i,k_i+1) , y\neq 0 \mbox{ or } x \in \{k_i, k_i+1\}\right\}
$$
since the points I have added to $A_i$ to form $B_i$ are in $M$. It is easy to see that the $B_i$ are path-connected (or you can use your argument about limit points to display $B_i$ as the union of two intersecting connected sets). The closed line segments in the set $M$ then join each $B_i$ to its east and west neighbours. To be even more formal, you can show by induction that $D$ can be constructed by repeatedly taking the union of pairs of intersecting connected sets.
