Choose some prime $p \neq 2,3$.
Now concatenate to the right side of it some prime $q<p$ so to arrive at some other prime.
Repeat until you cannot produce any more primes.
For example, if we take $p=5$ then we can proceed further to the prime $53$. Then we can proceed further to $5347$. Then to $5347103$. And so on and so on...
Intuition suggests that if we start with "big enough" primes that this cannot come to an end, that is, that always primes can be produced in this way. I did not check can we always proceed further if we start with all "small enough" primes.
Can we build in this way larger primes from smaller ones if we start with any prime $p \neq 2,3$ (I conjecture that the answer is yes) ? Intuitively the answer is: "of course we can", but how to prove this? What would be the key ingredients in the proof?
The question is whether, given any prime $p > 3$, there is a prime $q < p$ such that $10^d p + q$ is prime, where $q$ has $d$ decimal digits. I think it's highly likely that such $q$ always exists, but I doubt that it can be proven with current techniques.
You might look at OEIS sequence A066065. It is asserted there "a(k) < prime(k) for k > 2", but no proof or reference is given.