How to Prove that if $f(z)$ is entire, and $f(z+i) = f(z), f(z+1) = f(z)$, then $f(z)$ is constant?

I think you can just use Liouville's theorem. The thing is that $f$ is bounded on compact subsets of the complex plane, so in particular it is bounded in the unit square $S = \{ a + bi \in \mathbb{C} \mid a, b \in [0, 1] \}$.

So then you can use the two relations $f(z+1) = f(z)$ and $f(z+i) = f(z)$ to prove that the values of $f$ are determined by its values in the unit square because you can write $z = x + iy$ and then

$$f(z) = f(x + iy) = f(a + bi + (n + mi)) = f(a + bi)$$

where $0 \leq a, b \leq 1$ and $n, m \in \mathbb{Z}$. In this case $n = \lfloor x \rfloor$ and $m = \lfloor y \rfloor$. Then of course $f$ is bounded in the whole complex plane so Liouville's theorem applies, and $f$ has to be constant.

Here is a Hint: Liouville Theorem.