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?
Your proof is wrong. When you write
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 $\mathbb R^2$.