Prove or disprove that $A=(-1,1)$. 
Let $f:[0,1]\to [0,1]$ be a continuous and bijective function. Let $A = \{f(x)-f(y) : x,y\in [0,1]\backslash \mathbb{Q}\}.$ Prove or disprove that $A=(-1,1)$.

I know that there's no one-to-one function between the irrational numbers and the rational numbers since they have different cardinalities. Clearly $A \subseteq [-1,1].$ A continuous and one-to-one function is monotonic. So $\{f(0), f(1)\} = \{0,1\},$ and $1\not\in A.$ $0\in A$ and if $a\in A, -a\in A.$ Also f has the intermediate value property (i.e. it maps intervals to intervals). We can assume WLOG that $f$ is increasing as otherwise we can replace $f$ by $1-f$. Let $a\in (0,1)$ and choose $b\in [0,1]$ so that $f(b)=a$. For any $x\in (b,1)\backslash \mathbb{Q},$ by the bijectivity of f there's a unique $y_x$ so that $f(x)-f(y_x) = f(a)$ (since $f(x) > f(a)$). But how do I proceed from here?
 A: Since $f$ is bijective, continuous, and has a compact image, it must be a homeomorphism. Thus $f([0, 1] \setminus \Bbb{Q})$ must be a dense, $G_\delta$ subset of $[0, 1]$.
If we pick any $r \in [0, 1)$, then $f([0, 1] \setminus \Bbb{Q}) + r$ is a dense, $G_\delta$ subset of $[r, r + 1]$. Let
\begin{align*}
B_r &= f([0, 1] \setminus \Bbb{Q}) \cap [r, 1] \\
C_r &= (f([0, 1] \setminus \Bbb{Q}) + r) \cap [r, 1].
\end{align*}
Then, both $B_r$ and $C_r$ are dense, $G_\delta$ subsets of $[r, 1]$, and so there must exist a point $a$ in their intersection. Since $a \in B_r$, there exists some $x \in [0, 1] \setminus \Bbb{Q}$ such that $f(x) = a$. Since $a \in C_r$, there exists some $y \in [0, 1] \setminus \Bbb{Q}$ such that $f(y) + r = a$. Together, this implies $r = a - f(y) = f(x) - f(y)$, so $r \in A$.
A similar construction shows that, when $r \in (-1, 0)$, that $r \in A$ as well. As you showed, $\pm 1 \notin A$, so $A = (-1, 1)$.
EDIT Questions from the comments:

Why is $f([0, 1] \setminus \Bbb{Q})$ dense $G_\delta$?

Since $f$ is a homeomorphism, all the topological properties of sets are preserved, including density and $G_\delta$.
For density in particular, the inverse image of a dense set by a continuous map is known to be dense, and in this case, the continuous map is $f^{-1}$. So, the image of a dense set under our homeomorphism $f$ will be dense.
Note that $[0, 1] \setminus \Bbb{Q}$, being the complement of a countable set $[0, 1] \cap \Bbb{Q}$ in a complete space $[0, 1]$, is dense and $G_\delta$ (by the Baire Category Theorem), as we can write
$$[0, 1] \setminus \Bbb{Q} = \bigcap_{q \in \Bbb{Q}} ([0, 1] \setminus \{q\}),$$
which is a countable intersection of dense sets. Since $f$ is a homeomorphism (and hence a bijection),
$$f([0, 1] \setminus \Bbb{Q}) = f\left(\bigcap_{q \in \Bbb{Q}} ([0, 1] \setminus \{q\})\right) = \bigcap_{q \in \Bbb{Q}} (f([0, 1]) \setminus f(\{q\})) = \bigcap_{q \in \Bbb{Q}} ([0, 1] \setminus \{f(q)\}),$$
which is once again, dense and $G_\delta$.
Thinking about it, we could just shortcut the use of BCT and simply note that $B_r$ and $C_r$ have countable complements in $[r, 1]$, so their intersection must have countable complements, and so their intersection cannot be empty, as $[r, 1]$ is not countable.

How are $B_r$ and $C_r$ dense $G_\delta$ in $[r, 1]$?

As we established, $f([0, 1] \setminus \Bbb{Q})$ is dense and $G_\delta$ in $[0, 1]$. Since translating by $r$ is a homeomorphism between $[0, 1]$ and $[r, r + 1]$, this means that $f([0, 1] \setminus \Bbb{Q}) + r$ is dense and $G_\delta$ in the codomain $[r, r + 1]$. Since $0 \le r < 1$, we know that $[0, 1] \cap [r, r + 1] = [r, 1]$, where $[r, 1]$ has a dense (non-empty) interior.
We can write $f([0, 1] \setminus \Bbb{Q})$ as a countable intersection of dense open sets in $[0, 1]$, so intersecting each of these sets with $[r, 1]$ will produce open sets in $[r, 1]$. They will be dense in $[r, 1]$, since $[r, 1]$ has a dense interior. So, the resulting $B_r$ is dense and $G_\delta$ in $[r, 1]$ once more.
A similar argument works for $C_r$.

why must there be a point in their intersection?

The intersection of two $G_\delta$ sets is $G_\delta$. You form each by the intersection of countably many open sets, and the intersection is equivalent to intersecting over the union of these families. The finite (countable, even) union of countable sets is countable. Thus, the finite (countable, even) intersection of $G_\delta$ sets is $G_\delta$.
Because all of these open sets are dense in $[r, 1]$, Baire Category Theorem, once again, shows that their intersection, being a countable intersection of dense open sets, is dense. Thus, $B_r \cap C_r$ is dense and $G_\delta$ in $[r, 1]$. Note that if $B_r \cap C_r = \emptyset$, then this would contradict density!
