Proof verification regarding equal sets $RTP$ : { $12a+4b$ : $a,b ∈ $ Z} = { $4c$ : $c ∈ $ Z}
Solution : Let's first show that $LHS$ is a subset of our $RHS$. 
x ∈ { $x ∣4(3a+b)=x$ : a,b ∈  Z} → x ∈ { $x ∣4c=x$ : c ∈  Z}
Which is true. We can always choose $(3a+b)$ such that it equals $c$
Now let's show that $RHS$ is a subset of our $LHS$
x ∈ { $x ∣4c=x$ : c ∈  Z} → x ∈ { $x ∣4(3a+b)=x$ : a,b ∈  Z}
Which is true as well. We can always choose $c$ such that it equals $(3a+b)$. Correct?
 A: You want to actually find $a$ and $b$ such that $3a+b=c$, not state that they exist without proof. You are not choosing $c$ for the second part, as $c$ is already given.
A: In response to my request that you use more prose, you said, "we were asked to prove this statement clearly."  Proofs can be written very clearly in prose; and they can be written inscrutably in "symbolic" form.  I decided the best way to instruct you is by example.
Proposition  The sets $ A = \{ 12a + 4b | a,b \in \mathbb{Z} \} $ and $ B = \{ 4c | c \in \mathbb{Z} \} $ are equal.  
Proof  To prove this proposition, we need to show that $ A \subseteq B $ and $ B \subseteq A $.  
To show that $ A \subseteq B $, select any $ x \in A $.  By definition, we can find $ a $ and $ b $ in $ \mathbb{Z} $ such that $ x = 12a + 4b $.  Letting $ c = 3a+b $ we have $ x = 4c $ by the distributive property, whence $ x \in B $.  
To show that $ B \subseteq A $, select any $ x \in B $.  By definition there exists $ c \in \mathbb{Z} $ such that $ x = 4c $.  Let $ a = 0 $ and $ b = c $, and it is trivial to see that $ x = 4b $.  Thus, $ x \in A $, completing the proof.  $ \Box $
I hope I have clarified what I meant in my comments above.
A: Your proof is correct. Although, you should focus on the presentation of your proof.

$RTP$ : { $12a+4b$ : $a,b ∈ $ Z} = { $4c$ : $c ∈ $ Z}

How about labeling
$$A = \{ 12a+4b : a,b \in\mathbb  Z\}$$
$$C=\{ 4c : c \in\mathbb Z\}$$
and then starting the proof with we would like to show that $A=C$. Therefore, we need to show that $A\subseteq C$ and $C\subseteq A$.

Solution : Let's first show that $LHS$ is a subset of our $RHS$.
x ∈ { $x ∣4(3a+b)=x$ : a,b ∈  Z} → x ∈ { $x ∣4c=x$ : c ∈  Z}

Correct, assume that $x\in A$. Then $x=12a+4b$ for some $a,b\in\mathbb Z$. To show that $A\subseteq C$, you need to show that $x\in C$. As $x=12a+4b=4(3a+b)$, you have correctly deduced that $x\in C$ when $c= 3a+b$.

Now let's show that $RHS$ is a subset of our $LHS$
x ∈ { $x ∣4c=x$ : c ∈  Z} → x ∈ { $x ∣4(3a+b)=x$ : a,b ∈  Z}

Assume that $x\in C$. Then, $x=4c$ for some $c\in\mathbb Z$. To show that $C\subseteq A$, you need to show that $x\in A$. Let $a,b\in\mathbb Z$ be such that $a=0$ and $b=c$. Then, it is clear that $x=4b$. Therefore, $x\in A$.
You can end the proof with a sentence stating that as we have shown that $A\subseteq C$ and $C\subseteq A$, we have that $A=C$.
A: The proof is correct. In the second case, you can take $b=c$ and $a=0$ to make it more explicit. 
