If the top-left vertex was adjacent to all three of the right-side vertices, you would have $K_{3, 3}$, a bipartite graph. The nodes on the left form one partition, and the nodes on the right form another partition. Note that no nodes in a single partition are adjacent. So removing the three edges from the top-left vertex of $K_{3, 3}$ leaves you with your current graph.
So the graph you have is still bipartite, as you have:
-Two sets of vertices
-No pair of vertices in a given partition (set) are adjacent
Edit: Regarding your question on the maximum number of edges a bipartite graph on $n$ vertices can have without being connected. So we have one vertex disconnected. If $n$ is odd, then we have $K_{(n-1)/2, (n-1)/2}$. Otherwise, we have $K_{(n-1)/2, (n-1)/2 + 1}$. Then just count up the edges.
If the top-left vertex was adjacent to all three of the right-side vertices, you would have $K_{3, 3}$, a bipartite graph. The nodes on the left form one partition, and the nodes on the right form another partition. Note that no nodes in a single partition are adjacent. So removing the three edges from the top-left vertex of $K_{3, 3}$ leaves you with your current graph.
So the graph you have is still bipartite, as you have:
-Two sets of vertices
-No pair of vertices in a given partition (set) are adjacent
Edit: Regarding your question on the maximum number of edges a bipartite graph on $n$ vertices can have without being connected. So we have one vertex disconnected. If $n$ is odd, then we have $K_{(n-1)/2, (n-1)/2}$. Otherwise, we have $K_{(n-1)/2, (n-1)/2 + 1}$. Then just count up the edges.