Let $X$ be a metric space. Prove the equivalence of the following:
Please confirm if my attempt is fine or contains mistakes! Thank you so much!
My attempt:
(a) implies (b): Assume the contrary that $f:X \to \{0,1\}$ is a continuous surjection and $A = f^{-1}(0), B = f^{-1}(1)$. Then $A,B$ because $\{0\},\{1\}$ are open in $\mathbb R$ and $f$ is continuous. Moreover, $A \cap B = \emptyset, A \cup B = X$. Because $f$ is surjective, $A,B\neq \emptyset$. As such, $X$ is not connected, which is a contradiction.
(b) implies (a): Assume the contrary that $X$ is not connected. Then there are two nonempty open disjoint subsets $A,B$ such that $A \cap B = \emptyset, A \cup B = X$. We define $f:X \to \{0,1\}$ by $f(x)=0$ if $x\in A$ and 1 otherwise. It is easy to verify that $f$ is a continuous surjection, which is a contradiction.
"Then $A,B$" is not a sentence. You mean "Then $A,B$ are both open". Then the argument is correct, but don't say that $\{0\}$ and $\{1\}$ are not open in $\Bbb R$ but say they open in the induced subspace topology.
For the reverse actually show, instead of "claim", that $f$ is continuous. There is a helpful glueing lemma that might help. With that little gap filled, it's correct.