How do I prove that $X$ is connected? 
Consider the following subset of the plane $\Bbb R^2$:
  $$ X = \{x\times y\mid y=0\}\cup\{x\times y\mid x>0\text{ and } y=1/x\}. $$
  Prove that $X$ is not connected.

My attempt:
I think it has to do with these two sets form a separation in $X$.
Can anyone help me with this problem please?
 A: The function 
$$f(x,y):=xy$$ 
is continuous from $\mathbb{R}^2$, a fortiori $X$, to $\mathbb{R}$. Note that the topologies we consider on these sets are the ones induced by the usual Euclidean norms. Now
$$
f(X)=\{0,1\}
$$
is disconnected. Since the continuous image of a connected space is connected, $X$ can't be connected.
A: Show that both are closed in $\Bbb R^2$, and so closed in $X$. Hence, since they're pairwise disjoint (which you should be able to show), and their union is $X$, then they form a separation of $X$ (why?).
Showing that the first is closed, I leave to you. To see that the second is closed, note that the function $g:\Bbb R^{\ge0}\times\Bbb R\to\Bbb R$ given by $g(x,y)=xy$ is continuous, and that the second set is the $g$-preimage of the closed set $\{1\}$, so is relatively closed in the $\Bbb R^2$-closed set $\Bbb R^{\ge0}\times\Bbb R$, and so closed in $\Bbb R^2$.
A: HINT: Let $A=\{\langle x,y\rangle\in\Bbb R^2:y=0\}$, and let $B=\left\{\langle x,y\rangle\in\Bbb R^2:x>0\text{ and }y=\frac1x\right\}$. Clearly $X=A\cup B$.


*

*Prove that $A\cap B=\varnothing$.  

*Prove that $A$ and $B$ are both closed subsets of $X$. You can do this by proving that they are actually closed in $X$.  

*Explain why that makes $A$ and $B$ a disconnection of $X$.

