I need to prove the following result
Let $p$ and $q$ be 2 different primes. There isn´t any homomorphism from $\mathbb{Z}_p$ to $\mathbb{Z}_q$ or from $\mathbb{Z}_q$ to $\mathbb{Z}_p$
What I've done:
Suppose that exists an homomorphism $f : \mathbb{Z_q} \rightarrow \mathbb{Z_p} $ then $f (1) = 1$ so
$\underbrace{1+1+1\cdots+1\,}_\text{$q$ times} = \underbrace{f(1)+f(1)+f(1)\cdots +f(1)\,}_\text{$q$ times} = f(\underbrace{1+1+1\cdots +1\,}_\text{$q$ times}) = f(0) = 0$
So the left sum should be $0$ in $\mathbb{Z_p}$ but since $p \nmid q $ this is not true so we have a contradiction.
I think my proof isn´t correct because I've not used that $p$ and $q$ are primes, I've just used that they don´t divide to each other
For clarity, I'll denote by $0_k$ and $1_k$ the zero and unit element in $\mathbb{Z}_k$.
Your proof is actually correct. Indeed, you have $$ q1_p=qf(1_q)=f(q1_q)=f(0_q)=0_p $$ which is a contradiction, because it is known that $n1_k=0_k$ if and only if $k\mid n$.
You don't have used the “full force” of the hypothesis about $p$ and $q$ being primes and indeed it's not needed. The only fact needed in order to state that there is no ring homomorphism $\mathbb{Z}_k\to\mathbb{Z}_h$ is that $k\nmid h$, with exactly the same argument: $$ k1_h=kf(1_k)=f(k1_k)=f(0_k)=0_h $$ which is impossible if $h\nmid k$.
Using primality would allow for a different proof. The image of $f\colon\mathbb{Z}_q\to\mathbb{Z}_p$ is an additive subgroup of the codomain having order a divisor of $q$. Thus the image can only have order $1$, so it cannot be a subring.