I need to understand the following 

(i) $\{(x,y) \in R^2 |\;xy = 1\}\;U \{(x,y) \in R^2 |\;y = 0\}$ is not connected however 

(ii) $ Y = \{(x,y) \in R^2 |\;x^2 + y^2 < 1\}\;U \{(x,y) \in R^2 |\;y = 1\}$ is path connected. 

I attempt (i) along this line. For any $A = (a,0)$, I can find a ball of radius $1/2a$ such that $(a,1/a)$ lies outside this ball. Suppose $\exists (c,1/c) \in B(1/2a,A)$, clearly $(c-a)^2 + 1/c^2 \leq 1/4a^2.$ If I can contradict this, I am done. 

Noting that $a<c<a+1/2a$, $(c-a)^2 + 1/c^2 > 1/c^2 > 1/(a^2+1+1/4a^2)$. Now it suffices to show that $1/(a^2+1+1/4a^2) < 1/4a^2$, which is equivalent to showing $3a^2 > 1+ 1/4a^2$ which is true clearly $\forall$ a>1.

And here is my idea for path connectedness of (ii). To get a path from $(a,b)$ (interior of the circle) and $(c,d)$ (on the tangent), define $f:[0,1] \to Y$ to be $f(t) := t(a,b)+(1-t)(c,d).$ This is continuous and lies completely in Y as every point f(x) in Y except f(1) lies in the interior of the circle and f(1) lies in the tangent.

Are there any approaches for this that are more straight forward than this, say using some standard known results?