Can you divide $[0,1]$ into uncountably many disjoint sets? If you could do this, wouldn't it imply that $\mathbb Q$ is uncountable since each set would have a rational?
 A: You assume that each of the pieces of your partition contains a rational - why should this be true? It is true that every nonempty open set contains a rational, but there are lots of nonempty sets of reals which don't contain any rationals - e.g. the set of irrational numbers!
Indeed, $[0, 1]$ can be written as an uncountable union: $$[0, 1]=\bigcup_{x\in[0, 1]}\{x\}.$$ "Most" of these pieces are of the form $\{x\}$ for an irrational $x$, and so don't contain a rational.

More interesting partitions are possible. For $a, b\in [0, 1]$, let $a\sim b$ if $a-b$ is rational; then the resulting equivalence classes yield a partition of $[0, 1]$ into uncountably many countably infinite pieces $$[0, 1]=\bigcup_{x\in[0, 1]}\{y: x\sim y\}.$$ 
Indeed, we can even partition $[0, 1]$ into uncountably many uncountable sets: for $x, y\in[0, 1]$, write $x\approx y$ the $2n$th binary digit of $x$ and the $2n$th binary digit of $y$ agree for all $n$ (technically there's a bit of an issue here with reals with non-unique binary expansions - the dyadic fractions - but they're not hard to deal with).
Then the $\approx$-classes partition $[0, 1]$, and there are uncountably many of them, and each is uncountable.
A: You can do even more, you can write $[0,1]$ as a partition of an uncountable collection of uncountable sets.
Indeed we know that $[0,1]$ has the same cardinality as $[0,1]\times [0,1]$, therefore there is a bijection $\phi : [0,1]\times [0,1]\rightarrow [0,1]$.
Then the desired partition is $\bigcup _{a\in [0,1]} \phi ((a,[0,1]))$.
A: 
Wouldn't it imply that $\mathbb Q$ is uncountable since each set would have a rational?

No, since the statement each set would have a rational is wrong.
Some (a countable amount) would, and some (an uncountable amount) wouldn't.
Here is a list of disjoint sets whose union is $[0,1]$:
$$\forall{x\in[0,1]}:S_x=\{x\}$$
A: You cannot divide it into uncountably many disjoint intervals, since each must contain a rational as you note. However, it can be written as the uncountable union of singleton sets.
A: Yes you can: As others have already pointed out, 
$$ [0,1] = \bigcup_{x\in [0,1]} \{x\}$$
is an example. The Vitali set has been cited a few times as a partition of $[0, 1]$ into uncountably many countable sets. 
There are even partitions of $[0,1]$ into uncountably many uncountable sets. The family of sets $I_x = \{x\} \times [0,1]$, $x\in [0,1]$ is such a partition of $[0,1] \times [0,1]$. Fix a bijection $f\colon [0,1] \times [0,1] \to [0,1]$; then the family of images $(f[I_x])_{x\in [0,1]}$ is is an uncountable partition of $[0,1]$ into uncountable sets.
