Fix any $ \delta\in[0,1]$ and any $ \epsilon>0$. We have $ \prod_{p\ \mbox{is prime}} (1-\frac{1}{p}) = 0$ that is equivalent to \[ \sum_{p\ \mbox{is prime}} -\log(1 - \frac{1}{p}) = \infty.\] This is divergent series whose terms tend to 0. So, we can select a subsequence of primes $ p_1, p_2, \dots$ such that \[ \sum_{k=1}^{\infty} -\log(1 - \frac{1}{p_k}) = -\log\delta.\] Let $ m$ large enough integer such that \[ \left| \sum_{k=1}^m \log(1 - \frac{1}{p_k}) - \log\delta\right| < \min\{\frac{\epsilon}{2\delta},\frac{1}{2}\}.\] Since $ |\exp(x+t)-\exp(x)|<\exp(x)2|t|$ for $ |t|<\frac{1}{2}$, we have \[ \left| \prod_{k=1}^m \frac{p_k-1}{p_k} - \delta\right| < \epsilon.\] i.e., for $ n=\prod_{k=1}^m p_k$, we have \[ \left| \frac{\phi(n)}{n} - \delta\right| < \epsilon.\] Q.E.D.