Since I know nothing about Wang tiles, my notation and terminology are probably bad. I hope my argument is correct.

Let $W$ be a solvable finite set of Wang tiles.

A set $\mathcal A$ of finite blocks is *avoidable* (with respect to the fixet set $W$ of Wang tiles) if there is a ($W$-)tiling of the plane in which no element of $\mathcal A$ occurs.

$\emptyset$ is avoidable, since $W$ is solvable.

A set $\mathcal A$ is avoidable if and only if every finite subset of $\mathcal A$ is avoidable.

**Proof**. For the nontrivial direction, suppose every finite subset of $\mathcal A$ is avoidable. Since $W$ is finite, $W^{\mathbb Z\times\mathbb Z}$ is a compact space. ($W$ has the discrete topology, $W^{\mathbb Z\times\mathbb Z}$ the Tychonoff product topology.) $F_0=\{T\in W^{\mathbb Z\times\mathbb Z}:T\text{ is a proper tiling }\}$ is a closed set, as is $F_A=\{T\in W^{\mathbb Z\times\mathbb Z}:A\text{ does not occur in }T\}$ for each $A\in\mathcal A$. Since the family $\{F_0\}\cup\{F_A:A\in\mathcal A\}$ has the finite
intersection property, it has nonempty intersection; i.e., $\mathcal A$ is avoidable.

Hence there is a maximal avoidable set of finite blocks, choose one and call it $\mathcal A$.

Let $S$ be a tiling of the plane in which no element of $\mathcal A$ occurs. I claim that every finite block occurring in $S$ occurs infinitely often. Assume for a contradiction that some finite block $B$ occurs at least once in $S$ but occurs only a finite number of times. Then $S$ contains arbitrarily large squares in which $B$ does not occur at all, i.e., arbitrarily large squares can be tiled with no occurrence of any element of $\mathcal A\cup B$. Now another easy application of compactness shows that the whole plane can be tiled with no occurrence of any element of $\mathcal A\cup\{B\}$; that is, the set $\mathcal A\cup\{B\}$ is avoidable. But $B\notin\mathcal A$ (since $B$ occurs in $S$), so this contradicts the maximality of $\mathcal A$.

I guess this is the same idea as Aaron Meyerowitz's answer but more verbose. Well, maybe somebody will find a verbose answer easier to follow.

**P.S.** If your background is engineering and physics, you may not like my use of topological compactness. You can give an equivalent proof using König's infinity lemma, that a finitely-branching infinite tree has an infinite branch.