How to apply the substitution method to n/2? I recently was introduced to solving recurrence bounds by substitution but there's something i don't understand about it.

In standard induction proofs you prove a base case, assume it holds for n then show it holds for n+1 and therefore show it applies to integers larger than the base case.

For when you solve recurrences like T(n) = T(n/2) the proof has a base, assumes T(n/2) then proves it holds for T(n). Does this proof not only hold for the numbers that are multiples of two of the base cases? Would this not require an infinitely large number of base cases to show what it holds for?

Could someone please explain this to me?
You're entirely correct - if all you prove is $T(1)$ and $T(n/2) \Rightarrow T(n)$, then you prove that $T(n)$ holds for all powers of $2$. One way to fix that is to prove $T(n)$ for all odd $n$, but that defeats the idea of induction. Instead, usually one actually proves the stronger $T(\lfloor n/2 \rfloor) \Rightarrow T(n)$ or $T(\lceil n/2 \rceil) \Rightarrow T(n)$, and both are enough to deduce $T(n)$ for all $n$.

In many cases, if you can prove $T(n/2) \Rightarrow T(n)$ then you can also prove the stronger claims, but it's technically more awkward. So instead of dealing with this mess, you are only shown the weaker claim. The "interesting" ideas are found already there, and to prove the stronger claim requires just working harder but no new ideas (other than routine tricks used for handling floors and ceilings).