Two false equalities I have to prove that the following equalities are not valid.

First. Let $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$, $$[a,b)=\bigsqcup_{k=1}^n (a_k,b_k],$$
where $-\infty<a_k<b_k<+\infty$ for all $k\in\mathbb{N}$.

Proof.
If this were true there would be only one $k_0\in\{1,\dots, n\}$ such that $a\in (a_{k_0}, b_{k_0}]$ and therefore $a>a_{k_0}$. On the other hand, $(a_{k_0}, b_{k_0}]\subset [a, b)$, but $a_{k_0}\notin[a,b)$ since $a_{k_0}<a.$
Thanks to the suggestions I conclude that:
$\sup{[a,b)}=b$ and $\sup{\sqcup_{k=1}^n (a_k,b_k]}=b_n$ (We can assume without affecting the generality that $b_1< b_2<\dots< b_n$), then $b_n=b$ and therefore $b\in \sqcup_{k=1}^n (a_k,b_k]$, but $b\notin [a,b)$.

Second. Let $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$,
$$(a,b)=\bigsqcup_{k=1}^n(a_k,b_k],$$
where $-\infty<a_k<b_k<+\infty$ for all $k\in\mathbb{N}$.

Proof. If this were true there would be only one $k_0\in\{1,\dots, n\}$ such that $b\in (a_{k_0}, b_{k_0}]$, then $b\le b_{k_0}$. On the other hand we have tha $(a_{k_0}, b_{k_0}]\subset (a,b)$, but $b_{k_0}\notin (a.b)$. since $b_{k_0}\ge b$.
Are they formally correct? Thanks!

I do see your approach now now, but for the first, you would have to finish by saying that there is an $a'<a$ in $\bigsqcup_{k=1}^n (a_k,b_k]$; namely, $a' = (a_{k_0}+\min\{b_{k_0},a\})/2$ would do [so in particular, $a_{k_0} < a'<a$].
Your approach for your second does not work and I am not sure how one would fix. The problem is that why is there exactly one $k_0$ such that $b \in (a_{k_0},b_{k_0}]$.
How I'd approach both instead--and this proof does not assume disjointedness of the $(a_k,b_k]$s:
First note that $\bigsqcup_{k=1}^n (a_k,b_k]$ has a maximum element for any such choice of the $a_k$s and $b_k$s, whereas $(a,b)$ clearly does not have a maximum element and neither does $[a,b)$; indeed the maximum element of $\bigsqcup_{k=1}^n (a_k,b_k]$ is $\max\{b_1,b_2,\ldots, b_n\}$. So there is no way the first nor the second equation can be true for any such choice of the $a_k$s and $b_k$s.
