Are your tiles square shaped? One can then prove the result by what is essentially a compactness argument. Here is a brief idea:
Tile in order a square of size $1\times1$, then a larger square containing that one, of size $2\times 2$, then a larger one containing it, of size $3\times 3$, etc.
Suppose that your tiling allows us you to tile the plane in a non-periodic fashion. Then, for some $n$, you will have at least two options on how to tile the $n\times n$ square when you get there. Continue "on separate boards" with each of these two ways. Again, by non-periodicity, you should in each case reach a larger $m$ such that the $m\times m$ square can be tiled in at least two ways when you get there (of course, the $m$ in one case may be different from the $m$ in the other case). Continuing "on separate boards" in this fashion, you are building a complete binary tree, each path through which gives you a "different" tiling of the plane. The quotes are here, as we are not yet distinguishing between translations. But there are only countably many translates of a given tiling (if we insist that, say, the borders of our squares coincide with the lines of the form $x=n$ or $y=m$ for $n,m\in{\mathbb Z}$). But then, even after identifying translates, we are left with continuum many different tilings of the plane.
The other possibility is that your tiles does not allow non-periodic tilings. Then, no matter what path you follow, the process above should stop "splitting". But then you are only producing countably many tilings in this fashion.