Choose $n/2$ vertices and guarantee $3/4$ of edges are accounted for proof Give a polynomial-time algorithm that finds ceil(V/2) vertices that collectively
account for at least three-fourths (3/4) of the edges in an arbitrary undirected
graph.

The algorithm I have come up with is a greedy algorithm that iterates through the graph choosing the node with the largest number of incident edges and then updating the adjacent vertices with their new incident edge count.

I am fairly confident this algorithm will work, however I am struggling with the proof that I will always account 3/4 of the edges.

What I have so far is that because the sum of the degrees of all the edges in the graph is equal to 2E then on average every node will contribute at least a degree of 2E/n to that summation.  2E/n x n/2 vertices provides me with just E. Is this correct or am I missing something?
I'm not sure your algorithm works. Here is an algorithm that does work: choose $\lceil V/2 \rceil$ random vertices. You can calculate that each edge is chosen with probability at least $3/4$, and so the expected number of chosen edges is at least $3/4$ of all edges.

This algorithm can be derandomized (made deterministic) using the method of conditional expectations. I suggest you read about the method and work the details out. You will get something very similar to the greedy algorithm you're suggesting, but I suspect there will be some subtle differences.