functions - Is this a correct bijection between $(0,1]$ and $[0,1]$?

I need to give an explicit bijection between $(0, 1]$ and $[0,1]$ and I'm wondering if my bijection/proof is correct. Using the hint that was given, I constructed the following function $f: (0, 1] \to [0,1]$:
$$x \mapsto \left\{ \begin{array}{ll} 2 - 2^{-i} - 2^{-i-1} - x & \text{if } x \in (1-2^{-i}, 1-2^{-i-1}]\text{ for an } i \in \mathbb{N}_0 \\ 1 & \text{if } x = 1 \end{array}\right.$$

It's easy to see that for every $x \in (0, 1)$, there exists such an $i$.

Now define $\tilde{f}: [0,1] \mapsto (0,1]$ with
$$x \mapsto \left\{ \begin{array}{ll} 2 - 2^{-i} - 2^{-i-1} - x & \text{if } x \in [1-2^{-i}, 1-2^{-i-1})\text{ for an } i \in \mathbb{N}_0 \\ 1 & \text{if } x = 1 \end{array}\right.$$

I want to prove that $\tilde{f}(f(x)) = f(\tilde{f}(x)) = x$, so it has an inverse and therefore is a bijection. The case $x=1$ is trivial, so assume that $x \in (0,1)$ with $x \in (1-2^{-i}, 1-2^{-i-1}]$ for some $i \in \mathbb{N}_0$. This interval has length $1-2^{-i-1} - (1-2^{-i}) = 2^{-i-1}$, so we can write $x = 1-2^{-i} + \epsilon\cdot 2^{-i-1}$ for some $\epsilon \in (0, 1]$. We now calculate $f(x)$:
$$\begin{align*} f(x) &= 2 - 2^{-i} - 2^{-i-1} - x\\ &= 2 - 2^{-i} - 2^{-i-1} - (1-2^{-i} + \epsilon\cdot 2^{-i-1})\\ &= 1 - 2^{-i-1}(1+\epsilon). \end{align*}$$
We conclude that $f(x) \in [1-2^{-i}, 1-2^{-i-1})$. We now use the definition of $\tilde{f}$, so if we calculate $\tilde{f}(f(x))$, we get
$$\begin{align*} \tilde{f}(f(x)) &= 2 - 2^{-i} - 2^{-i-1} - f(x) \\ &= 2 - 2^{-i} - 2^{-i-1} - (2-2^{-i} - 2^{-i-1} - x) \\ &= x. \end{align*}$$

We conclude that $f$ has an inverse. Using exactly the same reasoning, we get that $f(\tilde{f}(x)) = x$ for all $x \in [0,1]$. Therefore it's inverse exists and it has to be a bijection.

I know there are less cumbersome methods of proving this fact, but as of now this is the only thing I can come up with.

Answer

It seems fine to me, I think, although it's a complicated enough construction that I'm not totally convinced of my surety.

If you want an easier method, by the way: let $f: (0, 1] \to [0, 1]$ by the following construction. Order the rationals in $(0, 1]$ as $q_1, q_2, \dots$. Then define $f(x)$ by "if $x$ is irrational, let $f(x) = x$; otherwise, let $f(q_i) = q_{i-1}$, and let $f(q_1) = 0$". We basically select some countable subset, and prepend 0 to it. You can do this with any countable subset: it doesn't have to be, or be a subset of, $\mathbb{Q} \cap (0, 1]$. If you prefer, for instance, you could take $\frac{1}{n}$ as the $q_n$.