Let G be a connected graph containing a cut-vertex $v$ and let $G_{1}$ be a component of $G-v$
(a) Show that the induced subgraph $G[V(G_{1}) \cup {v}]$ of $G$ is connected
(b)Show that the induced subgraph $G[V(G_{1}) \cup {v}]$ of $G$ need not be a block of G
To be honest I am not sure how to start the proof for this or which direction to take it in. Any help would be much appreciated.
I will help you with part (a). $G_1$ is a connected component of $G-v$, so by definition it is connected. Let's show that the induced subgraph $G[V(G_1)\cup v]$ is connected. It is enough to show that $v$ is connected to some vertex in $V(G_1)$. Since $v$ is a cut-vertex, there are vertices $x\in V(G_1)$ and $y\in V(G)\setminus V(G_1)$ such that there is a path from $x$ to $y$ passing through $v$ (this is a consequence of the fact that $G$ is connected and $v$ is a cut-vertex. try to show this by contradiction, for example). Let's denote the path by $P=xv_1v_2\dots v_r v v_{r+1}\dots v_n y$, then the vertices $\{v_1,v_2\dots, v_r\}$ are in $V(G_1)$ (because $x$ is in $G_1$ and $G_1$ is a component obtained after removing just $v$). In particular $v_r\in V(G_1)$, and this means that $v_rv$ is an edge in the induced graph $G[V(G_1)\cup v]$.
For part (b) you need to show that $G_1$ is not necessarily a maximal subgraph without cut-vertices. I think you can cook up an example.