Consider the following statements.
1.If every subset of a set is countable then set is countable.
2.If every proper subset of a set is countable then set is countable.
Then-
($a$)1 correct 2 may or may not correct.
($b$)2 correct 1 may or may not correct.
($c$)Both may or may not correct.
($d$)Both correct.
I attemped it using contrapositive of both the statements which are as-
(1) if a set is uncountable then there is atleast one subset which is uncountable.
(2) if a set is uncountable then there is atleast one proper subset which is uncountable.
My Argument:Both the contrapositive statements are satisfied by $\mathbb R$ which is uncountable but it has $\mathbb Z$ as subset which is countable.
From this it seems that (C) is true
But my argument seems not strong as "for example is not a proof".
Please clarify my doubt..
For part a, notice that every set is a subset of itself. For part b notice that removing even a single element gives a proper subset which is countable, so the original set is countable too.
Your example of $\mathbb R$ is not correct. The reason is that to falsify a "there exists atleast one" type of statement, it is not enough to give a counter-example. You have to show that such an example is impossible to construct.