I agree with Brad on the intended interpretation of the question. This means, as he said, that there is only one graph with no edges; if you think of the entire top row of your picture as one graph, that’s it. It also means that each of the three one-edge graphs in your second row is missing a vertex. The first, for instance, should be
a *--------* b * c
As Brad said, the three graphs in your bottom row are all the same graph, merely drawn differently. Thus, you should have a total of eight graphs, one with no edges, three with one edge each, three with two edges each, and one with three edges.
Now look at the hint in question (4), but apply it to these graphs: there are $3$ possible edges, $ab, bc$, and $ac$, and in any given graph on the vertex set $\{a,b,c\}$ each of these edges can be either present or absent. Thus, for each of the three edges you have a two-way choice when constructing a graph: keep the edge, or discard it. For instance, keeping $ab$ and discarding the other two yields the graph that I drew above. How many ways are there to make $3$ $2$-way choices? $2\cdot 2\cdot 2=2^3=8$, right? That’s why there are $8$ graphs on the vertex set $\{a,b,c\}$. Now try applying this logic to question (4).
Notice also that we could have counted the two-edge graphs on $\{a,b,c\}$ by asking how many ways there are to pick $2$ of the $3$ possible edges to keep. How many ways are there to pick $k$ things from a collection of $n$ different things? That idea should carry you through the rest of the questions.
Added in response to comments/questions: Yes, there are $\binom{n}2$ possible edges over $V_3$. Each pair of vertices is a possible edge, and from an $n$-element set you can make $\binom{n}2$ distinct pairs. To make a graph with vertex set $V_3$, you must choose to keep some subset of these $\binom{n}2$ possible edges. A collection of $\binom{n}2$ things has $2^{\binom{n}2}$ possible subsets, so there are $2^{\binom{n}2}$ possible graphs with vertex set $V_3$, not $2^n$. (This is the same mistake that you made in question (4).)
For question (8) you want to make graphs with $k$ edges. There are as many of these as there are ways to choose $k$ of the $\binom{n}2$ possible edges. This is not $\binom{n}k$; what is it?
I don't know what a "build-up" error is, but there is a minor, technical issue with your proof:
When proving the induction step, you should be starting from an arbitrary graph with $n$ vertices and $j + 1$ edges, and conclude from there. You do start with such a graph, but then fail to use it. Instead, you say
If we add an edge to the graph from the inductive hypothesis then we have...
To which graph are you adding an edge here, and what does it have to do with the graph with $j + 1$ edges? What you are doing is "building up" (is this a "build-up" error?) all the graphs you can from the inductive step, but failing to account for the possible graphs that cannot be built in this way!
Of course, every graph of $j + 1$ edges can be built up from a graph with $j$ edges: simply remove one edge. This is how you should be arguing. Start with a graph $G$ of $j + 1$ vertices, fix one edge $e$, and consider the graph $G'$ with $e$ removed. Then $G'$ has at least $n - j$ connected components. Adding $e$ back in to form $G$, by the arguments supplied, will remove at most one connected component, giving you at least $n - j - 1$ connected components.
(P.S. I don't want to contradict bounceback's comment; I too would award full marks, because your proof is almost 100% correct and would be doing better than 98% of my other students' efforts!)
Best Answer
Using degree sum formula: Choose a minimal counterexample such that $n$ edges connect $n+2$ vertices. Clearly, $n>1$. The graph can't have a leaf (contradicts minimality) and therefore, all degree have to be at least 2. But then, degree sum formula can't hold. So, there is no minimal counter example.
$\mathbf{Alternatively:}$ Let $G$ be a connected graph on $n+2$ vertices such that $e(G)=n$. Keep removing degree 1 vertices till all vertices are at least degree 2. Removing a degree one vertex might not give a graph with minimal degree 2. But if we keep to the process, we will eventually end up there. If we keep trimming and we end up at two vertices and an edge, then there weren't $n+2$ vertices to begin with. We can show that such a graph $G$ has a cycle. Start at some vertex $a_1$ and find $a_2 \notin \lbrace a_1\rbrace$ that it is connected to. Keep going till you run out of the vertices and have to come back to same $a_i$ that was already traversed. This process can only end if there is a dead end, a vertex of degree 1. But we removed them all. Now, remove an edge in this cycle and we can see that by taking the other route around, we still maintain connectivity in the graph. Again, trim all the vertices of degree 1 and get a new graph of minimum degree two. Find a cycle and remove an edge from it. Repeat this process over and over. It must bring the graph down to 0 edges but it will still have vertices left, by our assumption.