Prove the digital root of a square can be anything other than $2, 3, 5, 6, 8$? The digital root is the sum of the digits, unless that has more than one digit, so then you add up the digits again, until arriving at a single digit, e.g., $28$ -> $2 + 8 = 10$ -> $1 + 0 = 1$.
For what the digital root of a square can be, I just need to give an example of it, right? $0, 1, 4, 9$ are easy enough, and for $7$ I can just do $16$ or $25$ or $169$ or $196$ or $484$ or $529$ etc. (they seem to follow each other like that for some reason).
I'm not a professional mathematician, so please forgive me if what I'm saying is too obvious or too easy. I have looked among the first thousand squares and failed to find a single one with digital root $2, 3, 5, 6$ or $8$, but that doesn't prove anything, because maybe $1001^2$ does something contrary to expectation. I have this feeling that the explanation is in plain sight but for some reason I'm failing to see it.

First of all, $1001^2$ is not going to break the pattern you've observed so far, being $1002001$ and thus having a digital root of $4$. Squares with a digital root of $4$ follow or precede squares with a digital root of $1$, like $1000^2$.
But you're right in general to distrust the evidence of a thousand examples. In this particular case, the important thing to remember is that a problem of base $10$ digital roots is essentially a problem in arithmetic modulo $9$ (with just a couple of caveats to keep in mind), and modular arithmetic is periodic. Another neat feature of congruences is that you can multiply them and the results are valid, e.g., $2 \times 4 = 8$, $2 \times 5 = 1$, $2 \times 6 = 3$, etc.
So instead of trying to look at the infinitely many squares, you just need to look at the squares of $1, 2, 3, 4, 5, 6, 7, 8, (9)$, to get $1, 4, (9), 7, 7, (9), 4, 1, (9)$ (this also explains why the squares with digital root $7$ "seem to follow each other like that"). For the most part so far I've merely restated what others have said.
There is a slightly different way you can go about proving neither $3$ nor $6$ can be the digital root of a square in base $10$. In base $10$, we have these divisibility tests for $3$ and $9$: if the digital root if $3$, $6$ or $9$, then the number is divisible by $3$, and if the digital root is $9$, then the number is divisible by $9$. Then, if $n$ is a nonzero integer, then $3n$ has a digital root of $3$, $6$ or $9$. But $(3n)^2 = 9n^2$ and therefore it must have a digital root of $9$. So if a number $m$ has $3$ or $6$ for a digital root, that means it's divisible by $3$ but not by $9$, and $\sqrt{m} = x \sqrt{3}$. But $\sqrt{3}$ is not an integer. The only drawback of this method is that you can't easily extend it for $2, 5, 8$.
