Proving security of $\Pi' = E_k(E_k(m))$ knowing $\Pi = E_k(m)$ is secure I recently started studying cryptography but I am not sure I quite understand the concept of proof by reduction. Question I am trying to solve is as follow:
Suppose $\Pi$ is a symmetric encryption scheme which $ C \subseteq M$ ($M$ is message space and $C$ is ciphertext space). then we have $\Pi'$ with same key generation and decryption algorithm as $\Pi$ ($K' = K , D' = D$) with encryption algorithm as follow
$$E_k'(m) = E_k(E_k(m)).$$
I am trying to proof or reject:
a ) if $\Pi$ is indistinguishable in presence of an eavesdropper (simplest case when attacker can only see a ciphertext) then $\Pi'$ is indistinguishable.
b ) if $\Pi$ is CPA-secure then $\Pi'$ is CPA-scure.
for case a using proof by reduction i came up with solution

$ C \subseteq M $ implies a bijection between $M$ and $C$ so whenever $A'$ guess the chosen bit correctly $A$ will do as well so we have
$$Advantage\: of\:  A \geq Advantage\: of\:  A'$$
So if $A'$ be a attacker with non negligible advantage $A$ will be as well so a is true.
Am I using reduction correctly? What bout part b? Can we use almost same reasoning or there is an attacker for this case to prove $\Pi'$ is not CPA-secure?
EDIT :  regarding @Ievgeni answer for part a one time pad is a counterexample and regarding @Mikero comment i think reduction for part b might be something like below

conclusion :
part a : wrong . and one time pad is an counterexample and first picture is totally wrong.
part b : right . and proof is by reduction(picture two). for this reduction we have
$$Advantage\: of\:  A = Advantage\: of\:  A'$$
so if advantage of $A'$ be non-negligible advantage of $A$ will be too.
As Mikero noticed, the problem in your first proof is the fact that you can't suppose that $\mathcal{A}$ knows the secret $k$, and thus it can not encrypt the challenge.
And if you don't encrypt the challenge, then the input is not what is expected by $A'$. $A'$ is waiting $Enc^2_k(m)$ or $Enc^2_k(m')$ as a challenge not $Enc_k(m)$ or $Enc_k(m')$.
When you are doing a game-based proof, an important notion is the notion of
indistinguishability. For example If you are using an algorithm $\mathcal{A}$ as an oracle. And if you want to use some properties about the output of $\mathcal{A}$, it is important to check that the input of $\mathcal{A}$ follow the distribution which is mentioned in the property.
In your example, you are giving to the adversary $Enc(m)$ and not $Enc(Enc(m))$, it's completely different in the general case.
Let consider the xor encryption $Enc_k(m)= k\oplus m$.
It's really easy to build an attacker (let say a powerfull attacker who decrypts the ciphertext without the secret key) against $Enc^2$, it's just the identity function (Because $Dec_k^2=Enc^2_k$ is the identity function for all $k$.)
Then even $Enc$ is semantically secure, then $Enc^2$ could not to be. So a) is false.
But your proof about (b) seems correct to me.