I do not quite think that your approach can provide a contradiction.
A set of vertices, no two of which are joined to each other, is called an independent set. I will use this terminology. So our question is this : in a graph with hundred vertices and maximum degree four, we can find an independent set of size $20$.
Here's a better and easier way to do things. We will provide an algorithm that will find an independent set of size $20$ within the setup. Start with an empty set $S$.
Take any vertex $v$ in the graph, and add it to $S$.
Remove $v$ and its neighbours from the graph, and repeat the above step with the new graph.
End when there are no vertices left.
Now, what is the size of $S$? Note that at each step, at most five vertices removed at each step,since you remove the chosen vertex $v$ and its neighbours, of which there can't be more than four. Now, there are hundred vertices, which means that the steps given above repeat at least $20$ times. For each step, some new vertex is added to $S$. It follows that $S$ has at least $20$ elements.
Now, $S$ is an independent set. To see this, if two vertices $A$ and $B$ are in $S$ and are connected by an edge, then if (without loss of generality) $A$ was chosen before $B$ as a member of $S$ by the algorithm, then $B$ would have been removed from the graph, so it cannot have been made a member of $S$.
Consequently, $S$ is an independent set of size $20$.
Use the above approach for a generalization :
In every graph with $v$ vertices and maximum degree $d$, one can find an independent set of size $\frac n{d+1}$.
I am not sure that one can improve the number $\frac n{d+1}$. You can test this by trying trivial examples.
Best Answer
Each street crossing is a vertex of the graph. An avenue crosses about $200$ streets, and each of these crossings is a vertex, so each avenue contains about $200$ vertices. There are $15$ avenues, each of which contains about $200$ vertices, for a total of $15\cdot 200=3000$ vertices.
To make the description easier, imagine that the avenues all run north-south and the other streets east-west. Then each intersection of an avenue and a street has another intersection due north along the avenue. It also has one due south along the avenue, one due east along the cross street, and one due west along the cross street. That’s a total of four neighboring vertices. There must be an edge from the given vertex to each of those four, so we count $4\cdot 3000=12,000$ edges. But that’s counting each edge twice, once at each end, so there are really only half that many edges, or $6000$.
Now you might object that the vertex (i.e., intersection) in the northwest corner, say, has only two neighboring vertices, one to the east and one to the south, and similarly for the other three corners. You might also worry about the non-corner vertices along the edges, since they seem to have only three neighboring vertices each. But remember, the original figure of $200$ cross streets was only an approximation in the first place, so we might as well ignore these relatively minor edge effects: they probably don’t affect the result much more than the approximation in the $200$ figure already does.