Is the base case $m=0$ in this proof by induction correct? Let $\Bbb J(n)$ the set of bounded intervals in $\Bbb R^n$, that is, if $I\in\Bbb J(n)$ there are bounded intervals $I_k\in\Bbb R$ such that $I=\prod_{k=1}^n I_k$, where the product is the cartesian product (we consider the empty set an interval in $\Bbb R$ and consequently also in $\Bbb J(n)$).
Then let any finite collection $I,J_1,J_2,\ldots, J_m\in\Bbb J(n)$, and I want to prove using strong induction that
$$I\setminus\left(\bigcup_{k=1}^m J_k\right)\text{ can be written as a union of finite pairwise disjoint intervals}\tag1$$
If I choose the base case $m=1$ then the proof becomes messy, but if I choose the base case $m=0$ then I find that $(1)$ is just $I$, so the base case holds trivially, and the rest of the proof is very simple.

Proof: assume that a base case $m=0$ or $m=1$ holds, then we choose $(1)$ as hypothesis for all $1\le k\le m$ for some $m$. Then the induction step gives
$$I\setminus\left(\bigcup_{k=1}^{m+1}J_k\right)=\left(I\setminus\left(\bigcup_{k=1}^m J_k\right)\right)\setminus J_{m+1}\\=\left(\bigsqcup_{j=1}^r H_j\right)\setminus J_{m+1}=\left(\bigsqcup_{j=1}^r H_j\right)\cap J_{m+1}^\complement\\=\bigsqcup_{j=1}^r(H_j\cap J_{m+1}^\complement)=\bigsqcup_{j=1}^r(H_j\setminus J_{m+1})$$
for $H_j\in\Bbb J(n)$, and using again the hypothesis we can write $H_j\setminus J_{m+1}$ as a union of disjoint intervals, so we are done.

There is a huge difference in difficulty in this proof choosing $m=0$ or $m=1$ as the base case, so I'm not sure if the proof for $m=0$ is completely right. Can someone confirm that we can choose $m=0$ in this proof without harm?
Also, I would like to know if someone knows how to handle easily the base case $m=1$. I tried but I lost myself in a mountain of intersections, unions and abstract thinking.
Of course, if someone knows how to prove $(1)$ in a different way I would like to know it.

Equivalently $(1)$ can be stated as: for any finite collection $J_1,J_2,\ldots,J_m\in\Bbb J(n)$ then $\bigcup_{k=1}^m J_k$ can be written as a pairwise disjoint union of intervals.
However, with this assertion, the induction step is as hard as the base case of the above when $m=1$.

EDIT:
Amazingly it seems that the base case $m=0$ is valid. Observe what happens with this reformulation of $(1)$:

Let $I\in\Bbb J(n)$ and $\mathcal S\subset\Bbb J(n)$ finite. Then there is some finite $\mathcal B\subset\Bbb J(n)$ such that $$I\setminus\left(\bigcup\mathcal S\right)=\bigsqcup\mathcal B\tag2$$

Then from $(2)$ we can do induction over $|\mathcal S|$ and start with zero, so it seems that the base case of $m=0$ on $(1)$ is perfectly valid.
In the forum Rincón matemático professor Carlos Ivorra answered the question clearly so I will copy here a resume:

If $P(0)$ is true and $P(n)\implies P(n+1)$ then the proof by induction is not automatically correct (there is a classical example), we also need to check that $P(n)\implies P(n+1)$ can be shown true for all $n\ge 0$.

In this case we can't prove $P(0)\implies P(1)$ so the base case $m=0$ is not useful for a proof by induction.
In the forum Rincón matemático professor Carlos Ivorra answered the question clearly so I will copy here a resume:

If $P(0)$ is true and $P(n)\implies P(n+1)$ then the proof by induction is not automatically correct (there is a classical example), we also need to check that $P(n)\implies P(n+1)$ can be shown true for all $n\ge 0$.

In this case we can't prove $P(0)\implies P(1)$ so the base case $m=0$ is not useful for a proof by induction.