So I started with defining an Euler circuit as a closed walk containing at least one edge, not repeating any edge, and ending the walk on the same vertex as it was started. Is this a full proof, showing that an Euler circuit is connected by definition?
Or is there any way that is more legitimate to prove this?
The result is false as stated. For instance, let $G$ be a graph with vertices $v_1,v_2,v_3$, and $v_4$ and edges $e_1=\{v_1,v_2\},e_2=\{v_2,v_3\}$, and $e_3=\{v_3,v_1\}$. Then $G$ has an Eulerian circuit $v_1e_1v_2e_2v_3e_3v_1$ but is clearly not connected. An Euler circuit must include all of the edges of a graph, but there is no requirement that it traverse all of the vertices. What is true is that a graph with an Euler circuit is connected if and only if it has no isolated vertices: any walk is by definition connected, so the subgraph consisting of the edges and vertices making up the Euler circuit is connected, but there can be isolated vertices as well.