Throughout the post, I keep to the standard assumption that UFDs, PIDs, EDs and integral domains all refer to commutative rings. (But of course, there are noncommutative domains and PIDs and even some of the others, if you work hard enough ;) )
$R[X]$ is a UFD when $R$ is
For $B$, you can find in many commutative algebra texts that a commutative ring $R$ is a UFD iff $R[x]$ is. (For example, Corollary 16.20 in Isaacs Graduate Algebra)
Why is it clear, that a principal ideal domain is a integral domain?
Look back at your definitions: a principal ideal domain is just an integral domain with an extra property (having all ideals principal). A PID is a fortiori an integral domain.
After reading what you described about your definition, it sounds like maybe this didn't make it into your notes. A principal ideal ring is a ring in which all ideals are principal, but such a ring doesn't have to be a domain (For example, $\Bbb Z/\Bbb 4$ is a principal ideal ring, but not a domain, since $2^2=0$.) A (commutative) principal ideal domain is just a (commutative) principal ideal ring that is also a domain.
Hierarchy of properties
For $C$: Making a hierarchy like this is really good exercise. (In fact, I've embarked on pictures like that with dozens of ring types.) However, I hope you're not under the impression that you are going to organize all ring types linearly.
All of the domains you mentioned are subclasses of commutative rings, but the class of division rings is not contained in commutative rings. Out of all the rings you mentioned, there is one branch containing the domains:
$\text{field}\subseteq \text{Euclidean domain}\subseteq PID\subseteq UFD\subseteq\text{domain}\subseteq \text{commutative ring}\subseteq \text{ring}$
and then there is another branch
$\text{field}\subseteq\text{division ring}\subseteq\text{ring}$
You wrote that a PID "does not have a euclidean function" which is a bit like concluding that a rectangle does not have four equal side lengths. A PID does not necessarily have a euclidean function, but it might. Just like rectangles might have four equal sides, and hence be both squares and rectangles. You should just keep in mind that a Euclidean domain has more stringent structure than a PID, since they are a special subcase. Similar comments can be made about what you wrote about a UFD not having all ideals principal, etc.
$R[X]$ not a field
For $D$: To easily see that $R$ is not a division ring, just ask yourself if you can invert $X$ or not. When you multiply polynomials together, you're only going to get higher degrees of $X$. How will you get back down to $1$?
Inheritance
People have already pointed out how it's pretty easy to prove that $R$ is a domain iff $R[x]$ is, or the same for commutativity, and for the UFD property. Just in the last section we see that the case is not so for "being a field". Someone has also given an example that whlie $F[x]$ is a PID, $F[x][y]$ is not, so that property isn't preserved either. The same is also true for Euclidean domains since $F[x]$ is actually an example of a Euclidean domain.
Hopefully this will still be useful to you, or to someone in the future with a similar problem.
You have a number field $F$ and you want to find its ideal class group $C_F$. By Minkowski Bound Theorem every ideal class is represented by an ideal $I$ of norm $N(I) \leq c$ where $c$ is the Minkowski constant. So in order to find the elements of the class group, we need to find ideals of small norm in $O_F$.
There is a very important fact about ideals in rings of integers: $N(I) \in I$, so $I \mid (N(I))$. Now $N(I)$ is a natural number and can be factorised in the product of rational primes. So if we can factorise into primes all ideals $(p)$ with $p \leq c$, we will be able to find all ideals of small norm as their factors.
(Indeed prime and maximal ideals in Dedekind domains coincide.)
This is perhaps best illustrated by an example. Let $F=\mathbb{Q}(\sqrt{26})$. Then $O_F= \mathbb{Z}[\sqrt{26}]$, $n=2$, $r_2=0$ and $d_F=4\cdot 26 = 104$. The Minkowski bound is $c=\sqrt{26}<6$, so we need to find all prime ideals of norms $\leq 5$.
By Dedekind's Theorem for primes $2,3$ and $5$, we see that they factorise as
$(2) = (2, \sqrt{26})^2 =: P_2^2$ is a product of two prime ideals of norm $2$.
$(3)$ remains prime, so it has norm $9$, which is too large for our interest (ie the Minkowski Bound tells us that the same class is also represented by some ideal of smaller norm).
$(5)= (5, 1+\sqrt{26})(5, -1+\sqrt{26}) =: P_5 \cdot P_5'$ is a product of two distinct prime ideals of norm $5$.
Therefore, all ideals of norm $\leq 5$ in $O_F$ are $P_2, P_5$ and $P_5'$ and the ideal class group is generated by their classes $[P_2]$, $[P_5]$ and $[P_5']$. We have some relations between these already: $[P_5]$ and $[P_5']$ are inverses (because their product is a principal ideal) and $[P_2]$ has order $2$. Then we also observe that $(6-\sqrt{26})=P_2 \cdot P_5$ and so $[P_2] \cdot [P_5] = 1$ too and so $[P_2]=[P_5]=[P_5']$. After checking that the ideal $P_2$ is not principal, this means that the ideal class group has order $2$.
I hope this example was helpful and I'm happy to answer any further questions you may have.
Best Answer
I don't know if it's what you are looking for, but there is a thing called Hurwitz constant which can be easily computed from an integral basis of your number field. It is not very hard to show that it is an upper bound for a class number - it is certainly not a very good upper bound comparing to what you get from Minkowski, but you can easily work with it in case of a "small" ring and there is no big machinery behind that. For example, in the case of $\mathbb{Z}[\sqrt{-5}]$ you get a Hurwitz constant cca 10 and then it's not that hard to show that the class number is actually 2. I've seen this computation the other day and it took like two paragraphs.