the empty graph on $n$ vertices has $n$ components, each edge reduces the number of components by at most $1$.
Therefore we need to add at least $n-k$ edges, and as long as we don't form any cycles ( in other words only add edges between unconnected edges) $n-k$ edges will be enough.
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
Consider this problem combinatorially.
You have $n$ vertices partitioned into $k$ components. Assume $n \geq k$.
Consider the equation $n_1+n_2+\cdots+n_k=n$.
For each component with $n_i$ vertices, the maximum number of edges you can generate is $\displaystyle \binom{n_i}{2}$.
So, you want to maximize $\displaystyle \binom{n_1}{2}+\binom{n_2}{2}+\cdots \binom{n_k}{2}$.
Play with it for a while. The maximum result is obtained when $n_1=n-k+1$, and $n_2=n_3...=n_k=1$, and the maximum number of edges is $\displaystyle \binom{n-k+1}{2}$.