Decidable languages kleene star closure - question on a proof
It can be hard to grasp why we can just reason about an NTM to prove the Kleene star closure for decidable languages. The important fact is that any nondeterministic TM can be simulated by a deterministic TM. In some cases, it will just take a whole lot of time. When the topic is decidable languages, we are only concerned with whether something can be decided or not, regardless of how much time it will take to figure out. Therefore, an NTM will sometimes be more convenient to reason about, as we can abstract the "try all possible ways to do this", into "guess the right way to do this". However, the abstraction is only valid if we have a finite number of possibilities to try, of course.

So, for the Kleene star closure:

Assume that we have a decidable language $A$. Given that $A$ is decidable, there exists a decider for $A$, let us call it $M_A$. From this $M_A$, we can now construct a decider of $A^\ast$, let us call that decider $M_{A^\ast}$.

This is a description of how $M_{A^\ast}$ works:

On input $w$:

If $w = \varepsilon$ (the empty string), accept

For all possible splits of $w$ into $w_1,w_2,...,w_k$: if $M_A$ accepts all the strings $w_1, w_2, ..., w_k$, then accept

If all splits have been tried without success, reject.

Lets look at an example where $A = \{w \in \{a,b,c\}^\ast \; | \; w \; \text{starts with} \; ab \; \text{or ends with} \; c\}$

For instance, we have that $\{c, ab, ac, bc, cc, aac\} \subset A$, but $\{a, b, aa, ba, bb, ca\} \cap A = \emptyset$.

Now, let us look at how $M_{A^\ast}$ will determine whether or not the string $ccab$ is in $A^\ast$:

First, $M_{A^\ast}$ checks the trivial case. Since $w \neq \varepsilon$, it cannot accept right away. It now has to try all the 8 possible splits of $ccab$ listed here:

$S_1 = \{ccab\}$

$S_2 = \{cca, b\}$

$S_3 = \{cc, ab\}$

$S_4 = \{c, cab\}$

$S_5 = \{cc, a, b\}$

$S_6 = \{c, c, ab\}$

$S_7 = \{c, cab\}$

$S_8 = \{c, c, a, b\}$

You could argue that split $1$ is not a real split, but since it complies with our definition of a split, $M_A'$ has to check whether $A$ accepts all $w' \in S_1$, which is only $ccab$. $M_A$ rejects $ccab$ because it does not start with $ab$ or end with $c$.
In $S_2$, $cca$ is rejected by $M_A$, and hence $M_A'$ does not have to check $b$ and proceeds to split $S_3$.
In $S_3$, both strings are accepted by $M_A$. $cc$ is accepted because it ends with $c$, and $ab$ is accepted because it starts with $ab$. The machine $M_A'$ now accepts, and hence $ccab \in A^\ast$, simply because $cc \in A$ and $ab \in A$.