Proving a graph has no Hamiltonian cycle Show that $ G = (V, E)$ has no Hamiltonian cycle, where the vertices are $ V = \{ a, b, c, d, e, f, g \} $ and
the edges are $E = \{ ab, ac, ad, bc, cd, de, dg, df, ef, fg \}$.
From my working out, the vertices $ a, b, c, d, e, f$ are odd degrees of 3 and 1. Moreover $g $ being a even vertices of 2. 
There were three points that were made in my textbook to show that a graph does not contain a Hamilton circuit:


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*A graph with a vertex of degree one cannot have a Hamilton circuit.

*Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit.

*A Hamilton circuit cannot contain a smaller circuit within it.


According to the definition graph G does not have a Hamiltonian cycle because of the first definition. 
However, I am confused about 2 & 3 definitions and I am not sure if this graph involves them or not.  

As discussed in the comments, the three points are not definitions. They are just handy facts you can use to show that a graph is not Hamiltonian. If the facts don't apply to a given graph, it doesn't imply that it is Hamiltonian either - the test is just inconclusive.
Fortunately enough, we can use facts 2 and 3 to prove that the given graph indeed has no Hamiltonian cycle (note that fact 1 doesn't help us - $G$ has no leaf vertices). To do this:


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*Draw the graph with a blue pen, and label the degree of each vertex.

*Assume, towards a contradiction, that $G$ has some Hamiltonian cycle $C$.

*Apply fact 2 to each of the vertices of degree two. With a red pen, draw the edges that must be a part of $C$.

*Use fact 3 to get the desired contradiction.

