Show set of all differentiable functions $f\colon [0,1]\to\mathbb R$ is uncountable. Show set of all differentiable functions $f\colon [0,1]\to\mathbb R$ is uncountable. 
I think we have to use the Schröder-Bernstein theorem which states that if $|A| \leq |B|$,
and $|A| \geq |B|$ then $|A|=|B|$.
Need some help.
 A: To show that this set has the same cardinality as $\mathbb{R}$ proceed as follows.
For the one direction notice that your the set of differentiable function is a subsets of the continuous functions so their cardinality is less or equal their cardinality. 
A continuous functions is uniquely determined if we know the values it takes at the rationals.
Therefore they have the same cardinality as $\mathbb{R}^{\mathbb{Q}}$ but $$\mathbb{R}^{\mathbb{Q}}\cong \mathbb{R}^{\mathbb{N}}\cong {\left (\mathbb{2^\mathbb{N}}\right )}^\mathbb{N} \cong 2^{\mathbb{N} \times \mathbb{N}}\cong2^\mathbb{N} \cong \mathbb{R}. $$
Which proved that the continuous function have less or equal cardinality than $\mathbb{R}$.
A: Another way to look at this problem is to use the basics of metric spaces. As we know that continuous image of connected set is connected and $[0,1]$ is connected subset of $R$. Since,   differentiable maps from $[0,1]$ to $\mathbb{R}$  must be continuous though. Now, using continuous image of connected sets is connected it is clear that the set of all the continuous maps from $[0,1]$ to $\mathbb{R}$ must be connected subset of $R$ and The only connected subsets of R are the intervals (bounded or unbounded, open orclosed or neither) which must be uncountable.
A: Hint: For $c\in\mathbb R$ consider $f(x)=c$.
A: Take the collection of functions $f_\alpha(x) = \alpha$, with $\alpha \in [0,1]$. Since $[0,1]$ is uncountable, and each $f_\alpha$ is distinct, the set $\{ f_\alpha \}_{\alpha \in [0,1]}$ is uncountable.
Here $A=[0,1]$, $B= \{f:[0,1] \to \mathbb{R} | f \text{ is differentiable} \}$, and $\phi:A \to B$ given by $\phi(\alpha) = f_\alpha$ is injective.
A: Let $(f)_r:[0,1]\rightarrow \mathbb{R}$ be a family of functions over $\mathbb{R}$ defined as $f(x)=x^r$. Clearly differentiable, since exponential. Clearly uncountable since defined over $\mathbb{R}$
A: In a more general way, using @mathemagician idea, take any one differentiable function over $x$ and make it into a function over $ rx : r \in \mathbb R $ and obtain a continuum worth of differentiable functions through $f(x) \rightarrow f(rx): r \in \mathbb R $ each such $f(r_ox)$ is differentiable for all $r_o \in \mathbb R$ , as the composition of differentiable functions, and there are $\mathbb R=\mathbb c$ such functions.
You can e.g., take $  \sin x \rightarrow sin(rx) : r \in \mathbb R$ . Any one differentiable function will do.
You can see how to use a similar argument to show there are countably-many non-differentiable functions  $f: [0,1] \rightarrow \mathbb R$
