Why is $\{(x,y) \mid x,y\in\mathbb{C}\}$ an infinite dimensional vector space over $\mathbb{Q}$? $V=\{(x,y)\mid x,y \in \mathbb{C}\}$ is a vector space over $\mathbb{C}$, $\mathbb{R}$, and $\mathbb{Q}$. Then $\operatorname{dim} (V)=2$ over $\mathbb{C}$, and $4$ over $\mathbb{R}$, but why it is infinite over $\mathbb{Q}$?
Suppose that it were a finite-dimensional space of dimension $n \in \Bbb N$.
Let $B = \{v_1, \ldots, v_n\} \subset \Bbb C^2$ be a basis.
Then, every $v \in \Bbb C^2$ could be written as $$v = q_1z_1 + \cdots + q_nz_n$$ for some unique choice of scalars $q_1, \ldots, q_n$.
This gives us an injection $\Bbb C^2 \to \Bbb Q^n$ defined by
$$v \mapsto (q_1, \ldots, q_n).$$
This would give us a contradiction since $\Bbb C^2$ is uncountable whereas $\Bbb Q^n$ is not.

With a bit more effort, one can show that no countable basis exists either. For this, you may have to use the fact that the set of finite subsets of $\Bbb Q$ is countable.
You can see the dimension of a vector space $V$ over a field $\Bbb F$ as the number of parameters $a,b,c,... \in \Bbb F$ you need to use in order to write a generic element of $V$. Examples:
You can write every $(x,y)\in V$ as $(a,b)$ with $a,b\in \Bbb C$. $a$ and $b$ are 2 complex parameters so the dimension of $V$ over $\Bbb C$ is 2.
You can also write every complex number $a+bi$ with $a,b\in \Bbb R$. But to write every element $(x,y)\in V$ you need two complex numbers, so you can write it as $(a+bi, c+di)$ with $a,b,c,d\in \Bbb R$. You need 4 real parameters so the dimension of $V$ over $\Bbb R$ is 4.
How many rational parameters do you need to write every element of $V$?
If we say two and write $(x,y)\in V$ as $(a,b)$ with $a,b\in \Bbb Q$ we can find at least one element of $V$ that cannot be written by using just two rational numbers: $(π, i)$. Is this an element of $V$? Yes. But π is irrational so you need to sum an infinite number of rational parameters to represent it. (EDIT)&gt;&gt; So you must add it to a basis of $V$, so the dimension is three, at least. You can always find an irrational number with this same property and you'll have to add it to a basis of $V$. Since the linearly independent irrational numbers are infinitely many, the dimension of $V$ over $\Bbb Q$ is infinite.