Suppose \(\Gamma=(V,\set{~})\) is a totally disconnected graph, and let \(u_0,u_1\in V\text{.}\) Then \(u_0\) and \(u_1\) are (naturally) in separate components. The graph \(\Gamma'=(V,\set{u_0,u_1})\) contains no cycles, since a graph with only one edge cannot contain a cycle.
Suppose \(\Gamma=(V,E)\) is a forest and \(u_0,u_1\in V\) are vertices in different components of \(\Gamma\text{.}\) Assume, towards a contradiction, that \(\Gamma=(V,E\cup\set{u_0u_1})\) contains a cycle. Then there is a cycle \(C=(v_0,e_1,v_1,e_2,v_2,\ldots,v_{n-1},e_n,v_n=v_0)\) where \(v_i\in V\) and \(v_i\neq v_j\) when \(i\neq j\) and \(\set{i,j}\neq\set{0,n}\text{.}\) If \(u_0u_1\neq e_j\) for any \(j\in\set{1,2,\ldots,n}\text{,}\) then we have contradicted the assumption that \(\Gamma\) was a forest. On the other hand, if \(u_0u_1\in \set{e_1,e_2,\ldots,e_n}\text{,}\) then we may relabel the cycle \(C=(v_0=u_0, e_1, v_1, e_2, v_2, \ldots, v_{n-1}=u_1, e_n=u_0u_1, v_n=u_0=v_0)\text{.}\) But then \((v_0=u_0, e_1, v_1, e_2, v_2, \ldots, v_{n-1}=u_1)\) is a \(u_0,u_1\)-path, contradicting the assumption that \(u_0\) and \(u_1\) are in different components of \(\Gamma\text{.}\) In either case, we have a contradiction.