Let $p_n$ be the $n$-th prime ($n \geq 1$) and put $d_n = p_n - p_{n-1}$ for $n>1$ and $d_1 = 2$. Let $N\{S\}$ denote the number of elements in a finite set $S$. Show that for every $c>0$, $$N \left \{ n: p_n \leq x , d_n > c \log x \right \} \cdot c \log x < \sum_{p_n \leq x \\ d_n > c \log x} d_n \leq x$$ to deduce that for suitable $c,$ $$N \left \{ n: p_n \leq x , d_n > c \log x \right \} >> \pi (x).$$
I am not clear on how to show any of these statements. My idea for the last statement (not sure if it is the right approach) is to show $$N \left \{ n: p_n \leq x , d_n > c \log p_n \right \} < N \left \{ n: \sqrt{x} < p_n<x , d_n > c \log p_n \right \} + O(\sqrt{x}) < \frac{2x}{c\log x} + O(\sqrt{x}),$$ which I'm also having difficulty with. Any ideas on how to proceed/conclude?
Clarification:
$f(x) << g(x)$ if $\frac{|f(x)|}{g(x)} < M$ for $x$ sufficiently large.
$f(x) = O(g(x))$ if $\frac{|f(x)|}{g(x)} < M$ for $x$ sufficiently large.