(1) An improper integer greater than 1 has at least one improper prime divisor; and
(2) If $k$ is an integer coprime to $p$ and $n$ is a proper integer, then $kp - n$ is improper.
The first claim follows from the fact that the product of two proper integers is again proper. For the second, notice that $p$ does not divide $k n^{p-2}$ to deduce that
$$(kp - n)^{p-1} \equiv n^{p-1} - (p-1)k p n^{p-2} \equiv 1 + k p n^{p-2} \not\equiv 1,$$
so $kp - n$ is indeed improper (since $n$ is coprime to $p$, so is $kp - n$).
Since $\pm 1$ are both proper, letting $k \in \{1, 2\}$ and $n = \pm 1$ in (2) shows that $p \pm 1$ and $2p \pm 1$ are all improper, so each has an improper prime divisor by (1).

Since $p \ge 5$, the prime factors of $p \pm 1$ are all less than $p-1$; and since 2 is the highest common factor of $p-1$ and $p+1$, the conclusion follows, provided that 2 is proper.
Otherwise, look for an improper odd prime in the required range. To this end, notice that one of the numbers $2p \pm 1$ is divisible by 3, so its prime factors are all less than $p-1$, for $p \ge 5$; clearly, they are all odd, and the conclusion follows.

Alternative Solution.
If $p=5$, the primes $q=2$ and $r=3$ satisfy the required conditions, so let $p \ge 7$. In the setting of the previous solution, distinguish the following two cases:
If $p-2$ is improper, it has an improper prime divisor $q$ by (1). On the other hand, since 1 is proper, setting $k=n=1$ in (2) shows that $p-1$ is improper, so it has an improper prime divisor $r$ by (1). Clearly, $q$ and $r$ both lie in the required range, and they are distinct since $p-2$ and $p-1$ are coprime.
If $p-2$ is proper, set $k=1$ and $n = p-2$ in (2) to deduce that 2 is improper. On the other hand, since $(p-2)^2$ is proper, so is $-4p+4$. Setting $k=-3$ and $n = -4p+4$ in (2) shows $p-4$ improper, so it has an improper prime divisor $s$ by (1). Finally, since $p-4$ is odd, so is $s$, and it should now be clear that the primes 2 and $s$ satisfy the required conditions.