Connected set in $\Bbb{R}^2$ 
Which of the following sets are connected in $\Bbb{R}^2$ 
1) $A=\{(x,y): x^2y^2=1\}$
2) $B=\{(x,y): x^2+y^2=1\}$
3) $C=\{(x,y): 1<x^2+y^2<2\}$

My try:
From the shape of each set, $B$ and $C$ are connected whereas $A$ is not.My Question is how to prove it Mathematically?
For the set $A$, consider $f:\Bbb{R}^2 \rightarrow \Bbb{R}$ by $f(x,y)=xy$. Then $f$ is continuous and $$f(A)=\{f(x,y): (x,y) \in A\}=\{xy: x^2y^2=1\}=\{1,-1\}$$ 
If $A$ is connected, then so is its continuous image $\{1,-1\}$. The latter is not connected snd hence $A$ is not connected!
For the set $B$, define $g:\Bbb{R}^2 \setminus \{0\} \rightarrow S^{1}$  by $$g(x,y)=\frac{1}{x^2+y^2}(x,y)$$
Then $g$ is continuous and onto. Since continuous image of path-connected set is path- connected, $S^{1}$ is path-connected and hence connected!
Is my argument correct?
How to prove $C$ is connected  ?
 A: The set $B$ is the image of the function $$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb{R}^2\\&t&\mapsto&\bigl(\cos(t),\sin(t)\bigr)\end{array}$$and therefore, since $\mathbb R$ is connected and $f$ is continuous, $B$ is connected.
You can use the same argument for $C$. Consider the map$$\begin{array}{ccc}\mathbb{R}\times\left(1,\sqrt2\right)&\longrightarrow&\mathbb{R}^2\\(t,r)&\mapsto&\bigl(r\cos(t),r\sin(t)\bigr).\end{array}$$
A: $A$ is fine to show disconnectedness. You need a small argument that $f[A]= \{1,-1\}$: if $(x,y) \in A$ then $x^2y^2 = 1 = (xy)^2$, so that $xy = 1$ or $xy=-1$. This shows $f[A] \subseteq \{1, -1\}$. Both values are assumed as witnessed by $(1,1) \in A$ with $f$-value $1$ and $(-1,1)$ with value $-1$. This shows the reverse.
$B$ is fine as is. Maybe a small argument to show onto-ness? 
$C$ is connected, as a union of circles $x^2 + y^2 =c$, $c \in (1,2)$ "glued together" by $G=(1,2) \times \{0\}$ which is also connected (homomorphic to an open interval) and all circles intersect it. So we use it as glue to show connectedness of the union, via standard theorems on connected unions. Or use a continuous map as @JoseCarlosSantos does.
A: $A$ has two closed components, $ y=1/x$ and y=$-1/x$ so it is not connected.
$B$ and $C$ are path connected, so they are connected. 
