Is it obvious that $f\in \mathcal C^m([a,r])$ for all $rIn a book I'm reading, to prove that $f\in \mathcal C^m([a,b))$ they proved that $f\in \mathcal C^m([a,r])$ for all $r\in [a,b)$. 
Q1) Is it that obvious that $f\in \mathcal C^m([a,b))$ when  $f\in \mathcal C^m([a,r])$ for all $r\in [a,b)$ ? 
Q2) If $a<b-1$, can we "obviously" say that $$\mathcal C^m([a,b))=\bigcup_{n\in\mathbb N^*}\mathcal C^m\left(\left[a,b-\frac{1}{n}\right]\right) \ \ ?$$
Because unfortunately it's not that obvious for me, and I'm not really sure it's correct to be honnest.
 A: Q1) It is not true that if $f$ is once continuously differentiable, then it is $m$ times so. What their proof is doing is saying that you can prove the result with $m$ derivatives by repeatedly applying the one derivative case or that the general case is similar to the $m = 1$ case.
Q2) It is an intersection, not a union. (Remember, a union is a "or" and an intersection is an "and", so that the union of those sets is the set of functions that is continuous;y $m$ times differentiable on AT LEAST ONE $[a,b-1/n]$, but an intersection of those sets would require that it is on ALL those subintervals.) Second, you shouldn't require $a-b > 1$, but instead require that $n > 1/(b-a)$ so that $1/n < b-a$. 
Ask yourself, if a function is $m$ times continuously differentiable on $[a,b)$, then what does that mean? It means that they derivatives exist and are continuous at each point in $[a,b)$, because being continuously $m$ times differentiable is a local property: it only needs to be checked on some small interval containing the point. So, if a function is continuously $m$ times differentiable on $[a,b)$ then it is clearly continuously $m$ times differentiable on the sub intervals $[a,b-1/n]$ for $n$ large enough. Likewise, if it is differentiable on each of these subintervals, then for any point $x \in [a,b)$ there is an $n$ such that $x \in [a, b-1/n)$ on which $f$ is continuously $m$ times differentiable and so $f$ is continuously $m$ times differentiable near $x$. This means that $f$ is continuously $m$ times differentiable on the entirety of $[a,b)$.
A: 
Is it that obvious that $f\in C^m([a,b))$ when $f\in C^1([a,r])$ for all $r\in [a,b)$?

Yes, it is correct.
$f\in C^m([a,b))$ can be written as $f\in C^m([a,b-\varepsilon])$ for all $\varepsilon >0$ and $\varepsilon\leq b-a$ (to make sure right point of interval is bigger than left one).
Let $\varepsilon > 0$ and $\varepsilon\leq b-a$ be arbitrary but fixed. Then $a\leq b-\varepsilon<b$. Hence, we can set $r:=b-\varepsilon$. Since $f\in C^m([a,r])$ by assumption and $\varepsilon$ was arbitrary, it follows $f\in C^m([a,b))$.
Intuitively you can always choose $r$ as near to $b$ that it fits.

If $a<b-1$ can we “obviously” say that 
  $$C^m([a,b)) = \bigcup_{n\in \mathbb N^*}C^m\bigg(\bigg[a,b-\frac1 n\bigg]\bigg)\mathrm?$$

No, this is not correct. As @4-ier pointed out in his answer it has to be the intersection.
So
$$C^m([a,b)) = \bigcap_{n\in \mathbb N^*}C^m\bigg(\bigg[a,b-\frac1 n\bigg]\bigg) $$ is correct.
You need the assumption $a<b-1$ because if you set $n=1$ you get the interval $[a,b-1]$ where the right point has to be greater then the left one.
You can show the equality similar as for the above question. $\frac1 n\to 0$ for $n\to\infty$ but $\frac1 n > 0$ for all $n\in \mathbb N^*$. (Intuitively, this means you can approach $b$ as far as you want but you cannot reach it). Hence, you get the above equality.
