Show that for $p$ prime, $\mathbb{Z}_p$ is the only finite field with cardinality $p$ (not counting Isomorphism)
So here my proof. I know that the elements are $p$, in particular $\mathbb{Z}_p=\{ 0,1,2,...,p-1\}$. If I suppose there is another field $\mathbb{K}_p=\{ 0,1,2,...,p-1\}$ with $p$ elements, I want to show that $\mathbb{Z}_p==\mathbb{K}_p$.
An element mod $p$ in $\mathbb{Z}_p $ exist also in $\mathbb{K}_p$ so $\mathbb{Z}_p \subset \mathbb{K}_p$. But it's valid also the vice versa $\mathbb{Z}_p \supset \mathbb{K}_p$. Plus, every operation made using elements $z_i$ can fall into the $\mathbb{K}_p$ set and every operation made using elements $k_i$ can fall into the $\mathbb{Z}_p$ set. If those fields are finite, then must be $char(\mathbb{Z}_p) = p$ and $char(\mathbb{K}_p) = p$, so must be $\mathbb{Z}_p==\mathbb{K}_p$.
Are those evidence sufficient to prove the assertion?
I don't really understand your proof but I don't think that it's correct - you haven't really justified that every element of $\mathbb{Z}/(p)$ lies in $\mathbb{K}_p$, and then I'm not sure what your next sentence means. I think you have the right idea but haven't quite worded your proof correctly!
Here are the steps you should follow : suppose $K$ is a field with $p$ elements and multiplicative identity $1_K$. Let $\phi\,:\,\mathbb{Z}/(p) \rightarrow K$ be the ring homomorphism sending $1 \mapsto 1_K$, and show that this must be an isomorpsism