Finding Models for a sequent
You should be careful with order of quantifiers. You are given a predicate 
$$\exists x \exists y \forall z (z = x ∨ z = y)$$ 
which is read as 
$$\text{ there is } x \text { such that there is } y \text { for all } z \dots $$
while you read it as 
$$ \text { for all the values of }  z \text{ there exists } x \dots$$ 
which would correspond to 
$$\forall z  \exists x \dots$$
When you change the order of quantifiers the meaning (and hence the truth value) of the sentence changes. For example consider the set of natural numbers excluding 0. Then 
$$\forall x \exists y (x \text { divides } y)$$ is a true sentence since for any $x$ take $y=x$. But the sentence 
$$\exists x \forall y (x \text { divides } y)$$
is false since $x$ does not divide $x+1$.

Now as for your question, yes $A=\{0, 1\}$ models this sentence. In fact you could also take a set consisting of a single element, for instance $\{0\}$ or $\{1\}$ (there is no condition such as $x\neq y$).