A topological space such that every subspace from it is connected. 
Let $\tau$ be the topology in $\mathbb{R}^2$ whose open sets are of the form
  $$G_t = \{ (x,y)\in \mathbb{R}^2 \mid x>y+t\}, \; t\in \mathbb{R}$$
  Is the union of two lines $r\cup s$ connected in $(\mathbb{R}^2,\tau)$?

What I did was consider $r\cup s$ as subspace of $\mathbb{R}^2$ and assume that therexist $A$ and $B$ such that 
$$r\cup s = A\cup B$$
and $A\cap B =\emptyset$.
Then we can write $A$ and $B$ as
$$A=(r\cup s)\cap G_{t_1},\, B= (r\cup s)\cap G_{t_2}$$
But since for every $t_1, t_2 \in \mathbb{R}$, $G_{t_1}\subset G_{t_2}$ or $G_{t_2}\subset G_{t_1}$, we have that
$$A\subset B\; \text{or}\; B\subset A$$
In any case, $A\cap B\neq \emptyset$ and hence there is no separation of $r\cup s$, so $r\cup s$ is connected.
My question then is if this proof is okay, and also if it works too for any subspace of $(\mathbb{R}^2,\tau)$ since I haven't use at all anything about the two lines. Also

Are there other examples of topologies in $\mathbb{R}^2$ such that all subspaces are connected?


The indiscrete topology $\{\mathbb R ^2,\varnothing\}$ also works.  Or for a $T_1$ example, let $X:=\{\langle n,0\rangle:n\in\omega\}$ and take a basis of sets $U\subseteq \mathbb R ^2$ such that $U=V\cup (X\setminus F)$ for some open $V$ in $\mathbb R ^2$ and $F\subseteq X$ finite. You cannot get better than $T_1$ (if the space is Hausdorff, then any set of size $2$ is not connected).
By the way, the beginning space $\mathbb R ^2$ does not matter for these examples.
