I'd like to know if my proof is correct:
Assume $A$ is denumerable and and $A\subseteq X$ Assume X is uncountable (X uncountable means X is not finite and X is not denumerable)
$X=(X\backslash A) \cup (X \cap A)=(X\backslash A)\cup A$
(as $A \subseteq X$)
Therefore $X$ is uncountable $\implies (X\backslash A) \cup (X \cap A)$ uncountable.
Now using the Theorem: If A and B are denumerable sets then $A\cup B$ is denumerable. The contrapositive of this statement is: If $A\cup B$ is not denumerable then A is not denumerable or B is not denumerable.
Now applying the contrapositive to what we have so far
$(X\backslash A) \cup A)$ uncountable(not denumerable) $\implies(X\backslash A)$ uncountable or $A$ is uncountable.
(by the contrapositive of the theorem)
By assumption A is denumerable and therefore countable hence X\A is uncountable as required.