Robert Israel gives a nice answer using Tychonoff's theorem.
Here is an argument which uses diagonalization directly, avoiding Tychonoff's theorem.
Let the set of tiles be denoted $t_1,\ldots,t_K$. Let $\Delta$ be an upper bound to $diameter(t_k)$. By translation we may think of each $t_k$ as specific subset (polygon) in $\mathbb{R}^2$ containing the origin; let me refer to these specific tiles as the "prototiles". So in a general tiling, each of the tiles is a copy of some prototile $t_1,\ldots,t_K$ obtained by translation and rotation.
For each natural number $n$ let $\mathbb{T}(n)$ denote a finite tiling that covers $B(n+\Delta)$ (the closed ball centered on the origin of radius $n+\Delta$). Some tile in $\mathbb{T}_n$ contains the origin. So after translating $\mathbb{T_n}$ a distance at most $\Delta$ and then rotating $\mathbb{T_n}$ around the origin we may assume that $\mathbb{T_n}$ contains one of the protitles $t_1,\ldots,t_K$.
After these translations and rotations, although $\mathbb{T}(n)$ need no longer cover $B(n+\Delta)$, it still covers $B(n)$. In fact, $\mathbb{T}(n)$ has a finite subtiling that covers $B(n)$ and is contained in $B(n+\Delta)$, namely the union of all tiles in $\mathbb{T}(n)$ that intersect $B(n)$.
There are only finitely many of $t_1,\ldots,t_K$. So, there is a subsequence $n^0_1 < n^0_2 < n^0_3 < \cdots$ such that each of $\mathbb{T}(n^0_i)$ contain the same prototile.
In general the following statement is true for each natural number $n$:
$(*)_n$ There are only a finite number of tilings which contain a given prototile, which cover $B(n)$, and which are contained in $B(n+\Delta)$.
Using $(*)_1$, the sequence $(n^0_i)$ has a further subsequence $n^1_1 < n^1_2 < n^1_3 < \cdots$ such that each of $\mathbb{T}(n^1_i)$ has a common subtiling that covers $B(1)$ and is contained in $B(1+\Delta)$.
Using $(*)_2$, the sequence $(n^1_i)$ has a further subsequence $n^2_1 < n^2_2 < n^2_3 < \cdots$ such that each of $\mathbb{T}(n^2_i)$ has a common subtiling that covers $B(2)$ and is contained in $B(2+\Delta)$.
$\vdots$
By induction, we obtain an infinite nested sequence of subsequences $n^j_1 < n^j_2 < n^j_3 < \cdots$ such that for each $j$, each of the tilings $\mathbb{T}(n^j_i)$ has a common subtitling that covers $B(j)$ and is contained in $B(j+\Delta)$.
By diagonalization, we may extract a sequence $n(1) < n(2) < n(3) < \cdots$ which is simultaneously a subsequence of each of $n^j_1 < n^j_2 < n^j_3 < \cdots$. It follows that for each $m \ge 1$ the tilings $\mathbb{T}(n(m))$ all contain a common subtiling $\mathbb{T}_1$ that covers $B(1)$, for $m \ge 2$ they all contain a common subtiling $\mathbb{T}_2$ that covers $B(2)$, for $m \ge 3$ they all contain a common subtiling $\mathbb{T}_3$ that covers $B(3)$, ...
The union of the nested subtilings $\mathbb{T}_1 \subset \mathbb{T}_2 \subset \mathbb{T}_3 \subset \cdots$ is the desired tiling of the whole plane.
