I'm currently self-studying algebraic geometry and I tried to construct some proofs / counterexamples for relations between open / closed sets in Zariski and Standard Topology on $\mathbb{C}$ but I'm not sure about how to do the "open relations":
i) Closed in Zariski $\implies$ Closed in Standard This should be pretty clear. A closed set in Zariski is finite and any finite set is closed under Standard Topology.
ii) Closed in Standard $\implies$ Closed in Zariski Not true. Take the ball with radius 1 around the origin (with it's boundary) which is closed in Standard but isn't since it's cardinality is infinite.
Now what about the other implications, i.e. open in Zariski $\implies$ open in Standard and open in Standard $\implies$ Zariski? Do they just follow by De-Morgan laws from the above or are there any good counterexamples that show that these implications are right / wrong?
Are there any interesting connections between closed in Zariski and open in Standard or vice versa?
Your are right, every Zariski-closed set is also closed in the Euclidean topology, but the converse is false in general.
However, for complex algebraic group actions, the orbit is closed in Zariski topology if and only if it is closed in Euclidean topology, because orbits are constructible sets. For a proof see Borel's book on linear algebraic groups.
For more details see for example this post:
The Zariski closure of a constructible set is the same as the standard closure?