proof that BFS remains total after adding edge to graph
Let us assume that $G$ is a connected undirected finite graph.

Instead of by induction, a direct proof by contradiction as the following is probably easier and clearer.

Suppose we have just finished a BFS starting at vertex $s$. For the sake of contradiction, let us assume that there is a vertex $u$ of $G$ that is unvisited. Because $G$ is connected, there is a path from $s$ to $u$. Select the first unvisited vertex on that path and name it $u'$ (which may or may not be $u$). Let $s'$ be the vertex right before $u'$ on that path (which may or may not be $s$). Then $s'$ must be visited.

Now take a moment to reflect on BFS. Whenever a vertex is visited, each of it adjacent vertices must have been visited earlier or later; I will let you prove this proposition rigorously. However, this contradicts that $s'$, a visited vertex is adjacent to $u'$, an unvisited vertex. Our proof is done.

Here is the summary of the proof strategy. A subset of vertices of a connected graph that is closed under adjacency must contain all vertices. This also holds for infinite graphs.