The set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected Two disjoint sets $A$ and $B$, neither empty, are said to be mutually separated if neither contains a boundary point of the other. A set is disconnected if it is the union of separated subsets, and is called connected if it is not disconnected.
With the above definition of connected set, how to prove that the set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected.
Exercise should try using the definition of previous related and for this I guess I should do the proof by contradiction but do not know how to proceed.
 A: Do you know  straight lines are connected set in $\mathbb{R}^2$? do you know union of connected set is connected if they have one point in common?
Now can you see your set with above stated properties?
A: You can prove that $\{y=x\}$ and $\{y=-x\}$ are connected. Since they have a point in common, their union is. Can you try to argue this is the same, in essence, than showing the real line is connected?
First, let's obtain a slightly more useful equivalent to your definition. First, if $A,B$ are open sets and $A\cap B=\varnothing$, then they are separated. Now suppose $A,B$ are separated sets, and $C=A\cup B$. Then $C\setminus \bar A$ and $C\setminus \bar B$ are relatively open in $C$. But $$(A\cup B)\setminus \bar A=B$$ since $B\cap \bar A=\varnothing$. Thus $C=A\cup B$ and $A,B$ are relatively open in $C$. We can then just say that a set $C$ is disconnected if it can be written as the disjoint union of relatively open subsets of $C$. 
Now, suppose $f:C\to C'$ is continuous and onto. If $C'$ is disconnected by $A,B$, then $C$ is disconnected by the open sets $f^{-1}(A)$, $f^{-1}(B)$. Thus, by the contrapositive, if $C$ is connected, so must be $C'$. That is, the image of a connected set under a continuous function is connected.
Now consider the function $f:\Bbb R\to\Bbb R^2$ given by $f(x)=(x,x)$. Then your set is $f(\Bbb R)$. I hope you know that $\Bbb R$ is connected. You should see $f$ is continuous: $$\lVert f(x)-f(y)\rVert^2=2(x-y)^2$$
so if $x$ and $y$ are close, so is $(x,x)$ and $(y,y)$. Thus, $f(\Bbb R)$ is connected.
It is also true that the union of connected sets with non-empty interesction is connected. And one can prove it as follows. First, we prove that $C$ is connected if and only if the only continuous functions $f:C\to\{0,1\}$ $-$ where $\{0,1\}$ is given the discrete metric (topology), that is, all sets are open $-$ are constant.
If $C=A\cup B$ with $A,B$ open, then we can define $f(a)=1$ when $a\in A$ and $f(b)=0$ when $b\in B$, to get non constant continuous function $f:C\to\{0,1\}$. Conversely, if we had a  non-constant continuous function $f:C\to\{0,1\}$, we could write $C=f^{-1}(\{0\})\cup f^{-1}(\{1\})$ where this two sets are open and disjoint.
With this out of the way, suppose that we have a collection $\mathscr C$ of connected sets, and they have a point in common, call it $c$, that is $\bigcap\mathscr C\neq \varnothing$. Consider a continuous function $f:\bigcup \mathscr C\to \{0,1\}$. We will show that $f(x)=f(c)$ for any $x\in\bigcup\mathscr C$. Indeed, pick $x$ in the union. Then this belongs to some of the sets in the collection, call this set $C_x$. Then $f$ as a function from $C_x$ to $\{0,1\}$ is still continuous, so, since $C_x$ is connected, it must be constant. But $c\in C_x$; so we must have $f(x)=f(c)$. Since this $x$ was arbitrary, the claim follows.
A: If |x| < |y| then those x's are all in the disk of radius y about the origin.  That is surely connected
