Explicit homeomorphism between open and closed rational intervals? Sierpiński's theorem states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found here and a discussion here). 
An immediate corollary is that the rational intervals $[0,1],(0,1),[0,1)$ are homeomorphic. However, the theorem's proof is not very constructive and in fact quite complicated. (It's not a trivial theorem). 
Is there a way to construct an explicit homeomorphism between these intervals? 

It depends on how explicit you want to be. I’ll sketch one idea with $(0,1)$ and $[0,1)$; it can be extended to handle $[0,1]$ as well.
Let $x\in(0,1)$ be irrational, and for $n\in\Bbb N$ let $x_n=2^{-n}x$. (Note that $0\in\Bbb N$.) For $n\in\Bbb Z^+$ let $I_n=(x_n,x_{n-1})\cap\Bbb Q$, and let $I_0=(x_0,1)\cap\Bbb Q$. Each of the sets $I_n$ for $n\in\Bbb N$ is a clopen subset of $[0,1)$. The idea is to rearrange $[0,1)$, which looks like
$$0\,\ldots\,I_3\,I_2\,I_1\,I_0\;,$$
as $$I_1\,I_3\,I_5\,\ldots\,0\,\ldots\,I_4\,I_2\,I_0$$
and match that up with $(0,1)$. To do this, let $y\in\left(0,\frac12\right)$ be irrational. For $n\in\Bbb N$ let 
$$y_n=\frac12-2^{-n}y$$
and 
$$z_n=\frac12+2^{-n}y\;.$$
Finally, let $J_0=(z_0,1)\cap\Bbb Q$, $J_1=(0,y_0)\cap\Bbb Q$, and for $n\in\Bbb Z^+$ let $J_{2n}=(z_n,z_{n-1})\cap\Bbb Q$ and $J_{2n+1}=(y_{n-1},y_n)\cap\Bbb Q$. Each $J_n$ is a clopen subset of $(0,1)\cap\Bbb Q$, which looks like
$$J_1\,J_3\,J_5\,\ldots\,\frac12\,\ldots\,J_4\,J_2\,J_0\;.$$
For $n\in\Bbb N$ let $h_n:I_n\to J_n$ be a homeomorphism; then 
$$h:[0,1)\cap\Bbb Q\to(0,1)\cap\Bbb Q:q\mapsto\begin{cases}
\frac12,&\text{if }q=0\\
h_n(q),&\text{if }q\in I_n
\end{cases}$$
is a homeomorphism.
For the homeomorphisms $h_n$ you can use any order-isomorphism; these can be defined recursively in terms of an explicit enumeration of $\Bbb Q$. This probably isn’t as explicit as you’d really like, but I’m not sure that we can do much better.
