You cannot guarantee that $\langle a_n:n\in\Bbb N\rangle$ is a decreasing sequence: it might start out $$\left\langle 1,2,\frac1{10},\pi,\frac12,\frac13,\frac1{2^{100}},4,\frac1{100},\dots\right\rangle\;,$$ for instance. All you know is that all of its terms are positive, and for each $\epsilon>0$ there is some $n_a(\epsilon)\in\Bbb N$ such that $0<a_n<\epsilon$ whenever $n\ge n_a(\epsilon)$.
Similarly, all you can say for sure about $\langle b_n:n\in\Bbb N\rangle$ is that for every $\epsilon>0$ there is some $n_b(\epsilon)$ such that $1<b_n<1+\epsilon$ whenever $n\ge n_b(\epsilon)$. And above all you cannot assign specific values to the numbers $a_n$ and $b_n$: that’s changing the problem. (Of course, you can do so to look at an example or two in order to get a better idea of what’s going on, but that’s a different matter altogether.)
Now let’s take a look at $\liminf_n A_n$, where $A_n=[a_n,b_n)$: we want to determine which real numbers are eventually in the sets $A_n$, i.e., which are in all $A_n$’s from some point on. Here’s where those numbers $n_a(\epsilon)$ and $n_b(\epsilon)$ come in handy. For $\epsilon>0$ let $n(\epsilon)=\max\{n_a(\epsilon),n_b(\epsilon)\}$; then $0<a_n<\epsilon$ and $1<b_n<1+\epsilon$, and hence $$[\epsilon,1]\subseteq[a_n,b_n)\subseteq(0,1+\epsilon)\;.$$ for all $n\ge n(\epsilon)$. In other words, $[\epsilon,1]\subseteq A_n$ for all $n\ge n(\epsilon)$, and we conclude that for each $\epsilon>0$, $[\epsilon,1]\subseteq\liminf_n A_n$. It follows that $$\liminf_n A_n\supseteq\bigcup_{\epsilon>0}[\epsilon,1]=(0,1]\;.$$ On the other hand, if $x>1$, let $\epsilon=x-1$: for every $n\ge n(\epsilon)$ we have $1<b_n<1+\epsilon=x$, so $x\notin A_n$ whenever $n\ge n(\epsilon)$. This shows that $x$ isn’t even in infinitely many of the $A_n$’s, let alone in a tail of them, so $x\notin\limsup_n A_n$, and therefore certainly $x\notin\liminf_n A_n$. It’s also clear that no $x\le 0$ belongs to any of the $A_n$’s, so we’ve established that $\liminf_n A_n=(0,1]$, as you thought.
Along the way we’ve also seen that $\limsup_n A_n\subseteq(0,1]$, so $$\liminf_n A_n\subseteq\limsup_n A_n\subseteq(0,1]=\limsup_n A_n\;,$$ and it follows that $\limsup_n A_n=(0,1]$ as well, also as you thought.
It appears that you’re getting the concepts but might have a bit of difficulty actually writing down an argument to justify your reckoning of $\liminf$ or $\limsup$ of a sequence of sets.
We can "identify" each set with its characteristic function
$$\chi_A(x) = \begin{cases}1 &, x \in A\\ 0 &, x \notin A.\end{cases}$$
Then we have
$$\chi_{\liminf A_n}(x) = \liminf \chi_{A_n}(x); \quad \chi_{\limsup A_n}(x) = \limsup \chi_{A_n}(x)$$
for all $x\in X$. The characteristic function of the limes inferior/superior of the sequence of sets is the pointwise limes inferior/superior of the characteristic functions of the sets in the sequence.
Best Answer
The notation $\sup_n\{A_n\}$ is ambiguous, and I would avoid using it without more context. In the context of the $\limsup$ or $\liminf$ of sets, we are taking the partial order on sets by inclusion: $A \leq B$ if $A \subseteq B$. Then the supremum of a sequence $\{A_n\}_n$ is the smallest possible upper bound for every element of the sequence - the smallest set containing each $A_n$. This must be $$\sup_n A_n = \bigcup_n A_n.$$
Note that not every partially ordered set has well-defined suprema and infima - this kind of poset is called a lattice.
For a sequence of real numbers, the definition of limit superior is $${\limsup} \,\{a_n\}_n = \lim_{n \rightarrow \infty} \sup_{m \geq n} a_m.$$ The notation $\lim_{n \rightarrow \infty} A_n$ doesn't make sense for a sequence $\{A_n\}_n$ of sets. But observe that if $B_n = \bigcup_m A_{m \geq n}$ then $B_n$ is a nested sequence: $B_{n+1} \subseteq B_n$, since $B_{n+1}$ is the union over a smaller set of indices. Since $B_n$ is getting smaller and smaller, we might define the "limit" of $B_n$ to be small: the set of $x$ so contained in every $B_n$, or $\cap_n B_n$. Putting this together, a reasonable analog for the $\lim \sup$ of a sequence of real numbers applied to sets is
$$\limsup A_n := \bigcap_{n=1}^\infty \bigcup_{m =n}^\infty A_m.$$ Taking apart this definition, we see that $x \in \lim \sup A_n$ if and only if for all $n \in \mathbb{N}$ there exists $m \geq n$ so that $x \in A_m$: $$\forall n \in \mathbb{N}\, \exists m \in \mathbb{N} \,m \geq n \text{ and } x \in A_m.$$
This says that no matter how large of an $n$ you choose, I can find a larger $m$ so that $x \in A_m$. Another way of saying this is that there is no upper bound $n$ to the set of $m$ so that $x \in A_m$; and this is equivalent to saying that $x \in A_m$ for infinitely many $m$.
See the Wikipedia article for more info.