The flaw is when you use this sentence :
For every graph with n vertices and zero edges lets remove one vertice hence we get a graph with n−1 vertices and zero edges, by the assumpution the graph is connected, therefore the original graph is connected
for n = 2.
It doesn't work : you get 2 "1 vertice" graph, and nothing to tell about them. As in the all horses have the same color (twice "one horse has the same color as himself" doesn't proove that "2 horses have the same color").
For the proof to be correct, your base should be proven for $n$=2, not for $n$=1.
If you were able to prove the base for $n=2$, then you would be able to prove if for every integer :
Let suppose $P_2$ = "every (2-vertices,0 edge) graph is connected" is true, and then we can demonstrate $P_3$ = "every (3-vertices,0 edge) graph is connected"
Be $G$ a graph with 3 vertices ($V_1$ , $V_2$ and $V_3$) and zero edge. You can split $G$ in 3 graphes of 2 vertices (and zero edge each). By the assumption all those 3 graphes are connected (it's even more than needed). So there is a path from $V_1$ to $V_2$, a path from $V_2$ to $V_3$, and hence a path from $V_1$ to $V_3$ passing by $V_2$.
And so on...
But this kind of demonstration, you cannot do if you only suppose that $P_1$ = "every (1-vertice,0 edge) graph is connected" is true - you'll never be able to demonstrate $P_2$ = "every (2-vertices,0 edge) graph is connected" is true.
With just $P_1$ true :
Be $G$ a (2-vertices,0 edge) graph. By $P_1$ you can just say that each vertice is connected to itself , which is obvious and doesn't even need $P_1$ , but nothing more.