Let $G(V, E)$, a connected, undirected graph with positive weights. Let $E'\subseteq E$ such that the graph $G'(V, E') $ has no cycles. Find a spanning tree of $G$ which contains $E'$ edges, with minimum weight of all such trees (note: it doesn't have to be an MST)
My question is why can't I just find an MST and then make a union with $E'$? Id that's not this solution, what should I do instead?
If you make a union of the MST with $E'$ then the resulting subgraph might have cycles and, thus, be no tree. Therefore, your idea does not solve the problem.
Instead, you could start with $E'$ and then apply Prim's Algorithm.
To see the correctness of this approach, we modify your original problem. Since all edge from $E'$ have to be contained in the MST, we can remove each edge $e = (u, v) \in E'$ and identify both nodes $u$ and $v$ with each other. They will be connected by $E'$ anyway. Adding the same edges that Prim's Algorithm found in the original graph will give an MST for the reduced graph. Therefore, the union of these edges and $E'$ gives you a MST containing $E'$ in the original graph.