We have a measurable space $(\Omega, \mathbb{A})$. I want to prove the following statement about random vectors
$$X=(X_1, ..., X_k) \hbox{ is a random vector iff } X_i \hbox{ is a random variable } \forall i = 1,...,k $$
For this, I want to use only the following definition: $X:\Omega \to \mathbb{R}^k$ is a random vector (random variable if $k = 1$) if $[X \subset B] \in \mathbb{A}$, for all $B = B(q,r)= \{y \in \mathbb{R}^k : d_{max}(q,y) < r\}$, whith $(q,r) \in \mathbb{Q}^k \times \mathbb{Q}$. In other words, I want to use the fact that $\mathbb{R}^k$ and its topology have a countable base. In addition, we know that:
$$B(q,r) = \Pi_{i = 1}^k (q_i - r, q_i + r)$$ and
$$[X \in B(q,r)] = [q_1 - r < X_1 < q_1 + r, ..., q_k - r < X_k < q_k + r] = \bigcap_{i = 1}^k [q_i - r < X_i < q_i + r]$$
With this in hand, I would like to prove the most difficult implication: $(\implies)$. For this, take a randoms $x \in \mathbb{Q}^k$ and $r\in \mathbb{Q}$ and suppose that $[q_i - r < X_i < q_i + r] \notin \mathbb{A}$ for some $i$. Since that:
$$[X \in B(q,r)] \subset [q_i - r < X_i < q_i + r]$$
I want to prove the following fact: if $B \notin \mathbb{A}$ and $A \subset B$, $A \neq \emptyset$, then $A \notin \mathbb{A}$. With this, I can get some contradiction. Is it true?
Certainly your claim is false, as pointed out by Alessandro Codenotti. Use the fact that $B(q,r)$ can be written as countable union of sets of the form $[a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_k,b_k]$ to prove that $X$ is measurable if each $X_i$ is.