What's wrong with my "proof" that the Lebesgue measure of $[0,1]$ is $0$? 
Proposition. $\lambda [0,1] = 0$

Proof. Let $\varepsilon>0$ be arbitrary. We will prove that $\lambda[0,1] <\varepsilon$. Let $q : \mathbb{N} \rightarrow \mathbb{R}$ denote an injection with image equal to $\mathbb{Q} \cap [0,1]$. Let $p : \mathbb{N} \rightarrow \mathbb{R}_{>0}$ denote a sequence with $\sum_{i \in \mathbb{N}} p_i < \varepsilon$. Then $$\lambda [0,1] = \lambda \bigcup_{i \in \mathbb{N}} (q_i+[-p_i/2,p_i/2]) \leq \sum_{i \in \mathbb{N}} \lambda (q_i+[-p_i/2,p_i/2]) = \sum_{i\in \mathbb{N}} p_i < \varepsilon$$
So $\lambda[0,1] < \varepsilon$. Since this is true for all $\varepsilon>0$, we deduce that $$\lambda [0,1] = 0.$$

Question. What gives?


Yep. The fact that (in your notation)
$$
 [0,1] \subset \bigcup_{j=1}^{\infty} (q_j - p_j/2,q_j+p_j/2)
$$
is untrue is classic fact one has to get past to begin to understand Lebesgue measure. 
By picking $p_j = \epsilon/2^j$, you show exactly that there are open sets containing $\mathbb{Q} \cap [0,1]$ with Lebesgue outer measure at most $\epsilon$. (In particular they are measurable and hence the complement is measurable with measure $> 1 - \epsilon$, so they certainly don't cover $[0,1]$! ) 
