Using CRT to find arbitrarily long gaps between primes It is straight forward to find a gap between primes that consists of at least $2n$ using only the Chinese Remainder Theorem.
Let $p_n$ be the $n$th prime.
Find $x$ such that:
$$x \equiv -1 \pmod 2$$
$$x \equiv -2 \pmod 3$$
$$x \equiv -4 \pmod 5$$
$$x \equiv -6 \pmod 7$$
$$x \equiv -8 \pmod {p_5}$$
$$\dots$$
$$x \equiv -{2n} \pmod {p_{n+1}}$$
Clearly, not every condition is needed for the gap.  For example, $x\equiv -8 \pmod {3}$ so we could find a gap of at least $2n$ using CRT with less than $n+2$ distinct primes.
Limiting a set of equations such that every condition is needed for the gap size, it occurs to me that finding a prime gap of at least $2*p_{n}$ using CRT only need involve $n+2$ conditions each with distinct primes.
For example, a prime gap of at least $2*p_{5}=22$ can be found using the following $7$ conditions:
$$x \equiv -1 \pmod 2$$
$$x \equiv -2 \pmod 3$$
$$x \equiv -4 \pmod 7$$
$$x \equiv -6 \pmod 5$$
$$x \equiv -10 \pmod {11}$$
$$x \equiv -12 \pmod {13}$$
Using CRT, I find that $x = 9439$ which is prime and has a gap of $9461-9439 = 22$
To show how I came up with these set of conditions, let me use the following notation:
$$\#-\#-\#-\dots$$ 
to characterize the least prime factor of each odd integer in the gap which is not a prime.
In the above example, the gap of size $22$ consists of $10$ odd integers in the following order:
$$3-7-5-3-11-13-3-5-7-3$$
Finding a gap of $p_n$ consists of finding a set of $n+2$ CRT conditions that characterize this sequence.  For example, I can find a gap of at least $26$ using $8$ distinct conditions based on these $12$ odd integers:
$$11-3-7-5-3-13-17-3-5-7-3-11$$
It occurs to me that this approach will work for any $p_n$ using $n+2$ distinct primes as long as $2$ of the primes are greater or equal to $p_n$ and the order of the all but these two primes are the least prime factors in reverse natural order.  Natural order of the odd integers for the above would be:  $3,5,7,9,11$ with the least prime factors being $3,5,7,3,11$.  Reverse natural order of the least prime factors is therefore:  $11,3,7,5,3$.
Here's my question:
Does it follow that this method delivers the minimum number of distinct prime factors involved with a gap size of $2p_n$ or larger? 
It seems to me that the answer is yes. 
Let $f(m,p_n)$ be the number of distinct primes $p$ where $p < p_n$ and there exists $x$ such that $m < x \le (m+2*p_n)$ and $p$ is the least prime factor of $x$.
Does it follow that $f(m,p_n) \ge n+2$?  Is there a counter example where $f(m,p_n) < n+2$?

Edit 1:
I found a counter example for my second question.  In the case of $p_n=23$, $2*f_7(0,22) = 2$ but $f_7(76,122)=3$
Based on this, I am updating my question to just ask if there is an example of a prime gap that involve $n+1$ or less distinct least prime factors in the sequence of non-primes but is still $2p_n$ or larger.
 A: Your idea of using the least prime factors in "reverse natural order" is a good one which always works. To see this, let $x$ be the end point of this reverse order. What you've basically done is have, for each $1 \le i \le n$, the least prime $p_i$ be specified at $x - p_i$, which gives $x - p_i \equiv 0 \pmod{p_i} \implies x \equiv 0 \pmod{p_i}$. Combining these gives $x \equiv 0 \pmod{p_n\#}$. This covers not only the primes in reverse order back to $x - p_n$, but also the ones in forward order up to $x + p_n$. As such, each integer in $[x - p_n, x + p_n]$ has at least one prime factor associated with it except for $x - 1$ and $x + 1$, which you can use with any other $2$ primes, e.g., $p_{n+1}$ and $p_{n+2}$, giving a total of $n + 2$ primes being used.
You can extend this to $[x - (p_{n+1} - 1), x + (p_{n+1} - 1)]$, for a total prime gap size of $2p_{n+1}$ using $n + 2$ integers, which reducing $n$ by $1$ gives a gap size of $2p_{n}$ using $n + 1$ integers. This answers your basic question. You show this yourself with your example of a gap of $26 = 2 \times 13$, with $13 = p_6$ so $n = 6$. Among the $12$ odd integers, you only use primes up to $p_7 = 17$, so $7 = n + 1$ primes are needed. Thus, I assume your intent is to ask about having a gap size of at least $2p_n + 2$ using just $n + 1$ primes.
Note your strategy works well basically because you're trying to optimize the average number of integers each prime covers. Since there will be overlaps where an integer is covered by more than one prime, to get the most integers covered overall involves keeping the number of integers with prime overlaps to a minimum. One way to accomplish this quite well is to have one integer with almost all the primes overlapping, e.g., $x$ with your method.
However, this strategy is not always optimal. Using your notation, consider
$$\begin{equation}\begin{aligned}
3 - 5 - 11 - 3 - 13 - 17 - 3 - 7 - 19 \; - \\
3 - 23 - 5 - 3 - 11 - 7 - 3 - 5 - 13 - 3
\end{aligned}\end{equation}$$
Here $n = 8$ since there are $p_8 = 19$ odd integers covered, giving a guaranteed prime gap size of $2 \times 19 + 2 = 40$, but it only uses the $9 = n + 1$ primes up to $p_9 = 23$.
With your $f(m, p_n)$ function, you wrote it's

... the number of distinct primes $p$ where $p < p_n$ ...

However, this would mean $f(m, p_n) \lt n$, so your question asking if $f(m, p_n) \ge n + 2$ is always true doesn't make sense. Removing the restriction $p \lt p_n$, a counter-example to your question then is $f(0, 5) = 4$ since $5 = p_3$ giving $n = 3$, and the $0 \lt x \le 10$ range has just the $4 = n + 1$ least prime factors of $2$, $3$, $5$ and $7$.
