Overview
We'll introduce a sort of little hack called the missing modulo conjecture which can identify an integer's previous prime. We then show that this may not be perfectly reliable on account of how it interacts with all prime squares. In fact, we show that we would need a prime in every $(p^2,p(p+1))$ to be safe: we'd be forced to use them.
Flip that around, though, and it means that proving the MMC incidentally guarantees us that prime existence (which we do still need). So we prove it, and since we know it works, it follows that those primes must exist after all.
The missing modulo conjecture (MMC)
The gist here is that you can take any integer, look at the results you get from taking that integer mod every smaller integer, and determine the nearest prime by noting which small factors could not be generated.
For a given $n\in\mathbb N$, let $a$ be the smallest nonnegative integer such that $n \not\equiv a \pmod {m}$, where $m$ can take any value in the interval $[2,n/2]$. I assert that $n-a$ will always be the nearest prime $\leq n$. (See this question for more details.)
Prime squares
While the MMC mechanism most likely works perfectly, there's one big obstacle preventing its use in a proof.
Let $p$ be any prime. Then take $p(p+1) \not\equiv p \pmod{m}$ for $m\in \left[2,\frac{p(p+1)}{2}\right]$. This is the same operation as before aside from fixing $p$; it erroneously fails to return a factor of $p$, which is problematic because the typical range of $p$ in these cases could conceivably overlap or interfere with $a$, and here, $p$ does not point to a previous prime, meaning our $a$ values could also be suspect.
When $n=p(p+1)$, either there exists an $a<p$ and life is good, or we must default to $a=p$, since by definition we cannot have $a>p$ for the same $n$. If $a=p$, then since $n-a=p(p+1)-p=p^2 \not\in\mathbb P$, MMC would fail. For MMC to hold, we must always have our $a<p$.
However, this requires $n-a=p(p+1)-a$ be prime. And since $a<p$, this prime would need to be found within $\left(p^2,p(p+1)\right)$. Given this conflict arises for every prime $p$, if we want the MMC to bear out, we would need a prime to be located in every similar prime square interval. Conversely, if MMC were independently found to always hold, the existence of these primes would necessarily follow.
Solution: the MMC proof
Let there be some prime gap $q'<\ldots<q$, with $c=q-q'-1$ being the number of composite values in $(q',q)$. We're going to show that MMC works without issue in this instance, and by extension, everywhere else.
The lowest missing mod value $a$ on a prime $n$ is always $0$, and increments by $1$ with each higher $n$ until resetting to $0$ at the next higher prime. At $q-1$, we'll be up to $a=c$; this requires we establish that no values $0 \leq v < a$ are possible, where $v$ would be a hypothetical smaller missing mod result that we can't necessarily trust, like the prime square above.
$v<a$ iff $q-1 \not\equiv v \pmod {m\in[2,\frac{q-1}{2}]}$. In other words, we're making sure that any $v$ that exists is congruent to every last one of the $c$ possible lower values than $a$, or more precisely, making sure it would have had to be congruent$-$really, we're ruling out possibilities: if none of those values are missing from the results of our mods, then no $v$ using them could exist.
To check that, we establish that $q-k\equiv [0,a-k] \pmod {m}$, meaning that for each $k\leq c$, we want to map it to one of several $0$s we're about to find, at least one on each composite in the gap.
The lower bound of our range means that any starting-at-$0$ mod value in the range $[k+1,(q-k)/2]$ won't cycle around to $0$ until at least reaching $q$, so $k$ will be removed from the pool of could-have-been $v$ values. The best part is that because of the huge permissible range, you can simply rely on the greatest proper divisor of $q-k$ to qualify with no actual check or calculation required.
Repeat this until all choices for $v$ have been confirmed as observed, and therefore excluded. We'll have demonstrated that there cannot be a $v<a$, not even a bogus one like the prime square. With that addressed, there is no reason the same process cannot be implicitly applied to all prime gaps, which would prove MMC. And that in turn would prove there's always at least one prime in $\left(p^2,p(p+1)\right)$.
(example)
Consider the prime gap $(13,17)$. Identify three appropriate $k_i$ such that
$2\leq \gcd(k_1,16) \leq 8\qquad\qquad 3\leq\gcd(k_2,15) \leq 7\qquad \qquad 4\leq \gcd(k_3,14) \leq 7.$
We can go ahead and use $k_{1,2,3}=\{8,5,7\}$, the greatest proper divisors. These divisors correspond to mod values $m$ which would yield $0$, as they divide in cleanly: now, we know where some zeros are. Again, we're not changing anything; we're simply confirming things. Also, now we know that these are the mod values which are possible to receive within our parameters:
$ 14 \mapsto \{0,2,4\} \\ 15 \mapsto \{0,1,3\} \\ 16 \mapsto \{0,1,2,4\} $
Note the conspicuous absences of $\{1,2,3\}$, respectively. This is because those are the values $a$ takes when given each $n$. And note we have accomplished our goal of confirming the impossibility of $0$ for $14$, $\{0,1\}$ for $15$, and $\{0,1,2\}$ for $16$ as potential values for $v$. Collectively they comprise all values lower than our own $a=\{1,2,3\}$, meaning MMC definitely works for this gap interval.
Remember, this is only to illustrate the process, and because finding the divisors is guaranteed, the proof can be automatically applied to all gaps without any further work.
I'm looking forward to having someone confirm or point me at the mistake. Much thanks for either. Further updates/improvements to come.