This is an extract from Oksendal's SDE of the proof of the uniqueness of the solution of a SDE.
I cannot see how the $P[|X_t-\hat{X_t}|=0 \ \ \ \text{for all t} \in \mathbb{Q} \cap [0,T]]=1$ is derived. Also, I cannot see why is done just for the rationals.
Furthermore, I cannot see why how or why (5.2.11) is derived.
Because, if $f(t)$ is a continuous function, then the existence of any $t \in [0,1]$ such that $f(t) > \epsilon$ implies the existence of a rational $s \in [0,1]$ such that $f(s) > \epsilon$. In other words, for any $n \in \mathbb{N}$, the two sets $$\{\omega : \exists t \in [0,1] \text{ s.t. } |f(t,\omega)| > 1/n\} = \{\omega : \exists t \in [0,1] \cap \mathbb{Q} \text{ s.t. } |f(t,\omega)| > 1/n\}$$ are equal. To see this, call the set on the left $A_n$ and the set on the right $B_n$. Clearly $B_n \subset A_n$. For the converse, let $\omega \in A_n$ and for this $\omega$ let $t$ be such that $f(t,\omega) > 1/n$. Since $(n^{-1},\infty)$ is open, there is an open ball $U$ centred at $t$ such that $f(s,\omega) > 1/n$ for any $s \in U$. Pick a rational $s \in U$. Then $f(s,\omega) > 1/n$, so $\omega \in B_n$.
Since $P(A_n) = P(B_n) = 0$ for all $n$, the claim follows by taking complements and using subadditivity.