The closure of a connected set is connected. This is a proof by contradiction, unsing the definition of connectedness from Rudin, Principles of Mathematical Anaylsis. Can someone please check if it is any good?
Suppose S is a connected set, where its closure $ \bar{S} $ is not connected. Therefore there exist two nonempty sets A and B, such that $ \bar{A} \cap B = \bar{B}  \cap A = \emptyset $ and $ A \cup B = \bar{S}$. Define $ G:= A \cap S $ and $ H:= B \cap S $. Because G is a subset of A and H is a subset of B, it is clear that $ \bar{G} \cap H = \bar{H}  \cap G = \emptyset $. and $A\cup B = S$. So that S is not connected, contrary to our first assumption. q.e.d.
Thank you

The contradiction is that $G\cup H = S$ where G and H are separated, not $A\cup B=S$
