Is there a mistake in this proof? 
Proposition 1: The number of occurrences of an event within a unit of
  time has a Poisson distribution with parameter  $\lambda $ if and only
  if the time elapsed between two successive occurrences of the event
  has an exponential distribution with parameter  $\lambda $ and it is
  independent of previous occurrences.

Source
Now, if we look in the proof, it starts with:

Since  $X \ge x$ if and only if $\tau_1 + \tau_2 + \cdots + \tau_x \le
 1$ (convince yourself of this fact), the proposition is true if and
  only if:
$P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$ for any $x \in
 R_X$.

'The proposition is true if and only if' means that implication works in both ways. So we have two implications:
1) Proposition is true $\implies$ $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$
2) $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$ $\implies$ Proposition is true
The author doesn't say why two implications above are true. Could anyone explain that? It looks like the author proves the equality $P(X\ge x)=P(\tau_1 + \tau_2 + \cdots + \tau_x \le 1)$, but doesn't show why it implies Proposition 1. Therefore, the proof doesn't prove what it's claims to prove.
Note: the source has been updated and now takes into account the comments provided by the users on this page.

The $\text{“ if ''}$ assertion is correct; the $\text{“ only if ''}$ is not.
If the interarrival times are independent and exponentially distributed with rate $\lambda$ (and therefore with expected value $1/\lambda$), then the number of arrivals between time $0$ and time $1$ has a Poisson distribution with expected value $\lambda$.
But not $\text{“ only if ''}$.  For example, suppose with probability $1$ there are no arrivals during the second half of the interval of time, and during the first half we have waiting times until the next arrival that are independent and exponentially distributed with rate $2\lambda$ and expected value $1/(2\lambda)$.  Then we get a Poisson distribution with expected value $\lambda$ for the number of arrivals during that one unit of time.
However, we can get the conclusion about the exponential distribution of the interarrival times if we have a somewhat stronger hypothesis about the distributions of the numbers of arrivals.  Rather than supposing only that the number of arrivals during the whole time interval has a Poisson distribution with expected value $\lambda$, we further suppose that in every subinterval, the number of arrivals during that subinterval has a Poisson distribution with expected value equal to $\lambda$ times the length of the subinterval, and that the numbers of arrivals in disjoint subintervals are independent.  All that can make sense only if we know that the sum of independent Poisson-distributed random variables is Poisson-distributed.
Then we can show that the interarrival times are exponentially distributed and independent.  Here it is just for the first arrival:
$$
\Big[ (\text{time until the $x$th arrival}) > t \Big] \text{ if and only if } \Big[(\text{number of arrivals before time }t) \le x \Big]
$$
and consequently
\begin{align}
\Pr((\text{time until the 1st arrival}) > t) & = \Pr((\text{number of arrivals before time } t) = 0) \\[10pt]
& = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t},
\end{align}
so we have an exponential distribution of time until the first arrival.
