Bijection between $\mathbb{R}$ and $\mathcal{C}_0(\mathbb{R})$ It's possible to prove that $\mathcal{C}_0(\mathbb{R})$ (continuous function from $\mathbb{R}$ to $\mathbb{R}$)  and $\mathbb{R}$ have that same cardinality ('cause of you only have to choose the image for $\mathbb{Q}$).

But what's an explicit bijection?
Using Schroeder-Bernstein, you have that there are at least $|\mathbb R|$ continuous functions--the constant functions-- and there are at most $|\mathbb R^{\mathbb Q}|=|\mathbb R|$ functions; this last is the cardinality of the Real functions defined on the rationals , since not all functions defined on the dense subset of the rationals into the Realsextend (e.g., $f(x)= \frac {1}{(x-\ (2)^{1/2})})$ does not extend) into a continuous function from the Reals to themselves. I think Schroeder-Bernstein has an algorithm for producing a bijection.
Since $\mathcal{C}_0$ contains continuous functions, each function can uniquely express as a Taylor Series which is a polynomial of infinity degrees. You put all the coefficients of this polynomial together with the constant term and you can create an ordered sequence which is unique. eg, for $f(x)=\sin(x)$, you can express as $f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...=x-\frac16x^3+\frac1{120}x^5...$. Then the sequence you create is $&lt;0,1,0,-\frac16,0,\frac1{120},......&gt;$

Let $\mathcal S$ be the set of all these function, then $\mathcal C_0=\mathcal S\preceq\mathbb R^{\mathbb N}=\mathbb R$. Of course $\mathbb R\preceq \mathcal C_0$ since for each real number, there is a constant function for it. Then by Schroeder Bernstein Thm, $\mathbb R\sim \mathcal C_0$