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I have managed to write down two proofs showing the connectedness of $K$. But still shaky about both of them. Here are the proofs:

1)Suppose $K$ is disconnected. Then we write it's separation as $ K = U \cup V $. Hence $ K $ is open since U and V are open.

Now, since each $K_i$ is compact in $\Bbb R^2$, hence by Heine-Borel theorem it is closed and bounded. Hence $K$ is also closed and bounded and non-empty.

We have got that $K$ is open as well as closed. But the only open and closed sets in $\Bbb R^2$ are ${\emptyset}$ and the whole $\Bbb R^2$. But because $K$ is non-empty and bounded, hence it is not one of them. Hence our assumption was wrong. Hence K must be connected.

another proof I thought was like this:

2)Let $f : K_1 \to \{0,1\} $ be a continuous function. Then it's constant by connectedness of $K_1$ . Say $f(K_1)=0$. Then restricting the function to K, we get that $f(K)=0$.

Can any other proof be given? are these proofs valid?

Error 404
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Your proof is wrong. When you write

1)Suppose K is disconnected. Then we write it's separation as K=U∪V. Hence K is open since U and V are open.

you've gone off the rails. For $K$ is disconnected if there are open sets $U$ and $V$ in $K$'s topology with the property that you specify. But what does it mean for $U$ to be open in $K$? It means that $U$ is $K \cap U'$, where $U'$ is an open set in $\mathbb R^2$. For instance, the set $X$ consisting of the two points $(0, 0)$ and $(1, 0)$ constitute a disconnected set in $\mathbb R^2$, but if you were to write a disconnection of this set, you'd have to have $U = \{(0,0)\}$ and $V = \{(0, 1)\}$, neither of which is an open set in $\mathbb R^2$, and as you'll note, $X$ is not open in $\mathbb R^2$ either. Of course, $X$ is open as a subset of $X$ (in the inherited topology), but that doesn't make it open in $\mathbb R^2$.

John Hughes
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    proof 2 is also wrong; it is implicitly assuming that any continous function on $K$ can be lifted to a continuous function on $K_1$, which is false. – hunter May 16 '15 at 13:54
  • So how do we write then the separation of the set containing $(0,0)$ and $(1,0)$? Then why do they say that a separation is disjoint union of two non-empty OPEN sets in $X$ for $X$ being the main big set containing the set which is under consideration of connectedness? – Error 404 May 16 '15 at 13:59
  • The separation of $X$ by $U$ and $V$ is correct, but for $U$ to be open in $X$ means that $U$ is the intersection $X \cap U'$ of $X$ with some set $U'$ that's open in $\mathbb R^2$, not that $U$ itself is open in $\mathbb R^2$. In this case, we can take $U'$ to be an open ball in $\mathbb R^2$ of radius $\frac{1}{2}$ about $(0,0)$, and $V'$ to be an open ball of radius $\frac{1}{2}$ about $(1,0)$. – John Hughes May 16 '15 at 15:00
  • Agreed to you @JohnHughes. But my U and V are open in R^2. – Error 404 May 16 '15 at 15:07
  • I guess I'l have to go again through the definition of connectedness. I am sorry. – Error 404 May 16 '15 at 15:10
  • I think that carefully reviewing the definition of connectedness and of the relative topology would be a good idea. Best of luck! – John Hughes May 16 '15 at 15:29