Questions about the induction on cardinals From Hereditary Cardinality and Rank :

For an infinite cardinal $\kappa$, $$\forall x,\ \textrm{hcard
 }x<\kappa\rightarrow\textrm{rank }x<\kappa$$


We can show this by induction on $\kappa$. Suppose $\kappa$ is a
successor cardinal; that is, $\kappa = \lambda^+$ for some $\lambda$.
Then, if $hcard(x)<\kappa$, $|x|<\kappa$ and, by the induction
hypothesis, $rank(y)<\kappa$, for $y\in x$. Since $\kappa$ is regular,
it follows that $rank(x) <\kappa$. Now, suppose that $\kappa$ is a
limit cardinal. Then, if $hcard(x)<\kappa$, there is $\kappa'<\kappa$
such that $hcard(x)<\kappa'$ and thus, by the induction hypothesis,
$rank(x)<\kappa'<\kappa$.

I'm only familiar with the induction on ordinals, but I'm assuming the induction on cardinals is very similar. However, I still don't get this proof. My understanding is that we should prove two things:

if  $$\forall x,\ \textrm{hcard  }x<\kappa\rightarrow\textrm{rank
 }x<\kappa$$
then
$$\forall x,\ \textrm{hcard  }x<\kappa^{+}\rightarrow\textrm{rank
 }x<\kappa^{+}$$

and

if $\lambda$ is a limit cardinal and for all  $\kappa < \lambda$ we
have  $$\forall x,\ \textrm{hcard  }x<\kappa\rightarrow\textrm{rank
 }x<\kappa$$
then
$$\forall x,\ \textrm{hcard  }x < \lambda \rightarrow\textrm{rank
 }x<\lambda $$

Is this general outline correct? Also, is there some kind of "base of induction" case that needs to be considered?
When proving the statement for the successor cardinal, the author states: "by the induction
hypothesis, $rank(y)<\kappa$, for $y\in x$". Where is this assumption coming from? Further, how does the regularity of $\kappa$ imply that $rank(x) < \kappa $?
The case of limit cardinal is much clearer (at least, I can see how the assummption that I stated above is used). The only unclear thing here is why there is $\kappa' < \kappa$ such that $hcard(x) < \kappa'$. The fact that $\kappa$ is a limit ordinal means that for any cardinal $\kappa_1 < \kappa$ there is a cardinal $\kappa_2$ with $\kappa_1 < \kappa_ 2 < \kappa$. But why does $hcard(x)$ have to do with $\kappa_2$ (or $\kappa_1$)?
 A: 
When proving the statement for the successor cardinal, the author states: "by the induction
hypothesis, $rank(y)<\kappa$, for $y\in x$". Where is this assumption coming from?

I am also confused by this part of the proof, and suspect it isn't quite correct. It seems like something very similar works if you do it instead as a proof by $\in$-induction:

Let $\kappa$ be a regular cardinal. Then $|\operatorname{trcl}(x)| <
\kappa\to \operatorname{rank}(x)<\kappa.$
Proof. By $\in$-induction on $x.$ Let $|\operatorname{trcl}(x)|<\kappa.$ If $y\in x,$ then $|\operatorname{trcl}(y)|<\kappa,$ so by the induction hypothesis, $\operatorname{rank}(y)< \kappa.$ Then $\operatorname{rank}(x) = \sup_{y\in x}\{\operatorname{rank}(y)+1\}$ which is less than $\kappa$ since $\kappa$ is regular and $|x|<\kappa.$

Then you can improve it any cardinal by noting that if $\kappa$ is singular and $|\operatorname{trcl}(x)|<\kappa,$ then $|\operatorname{trcl}(x)|<\lambda$ for some regular $\lambda < \kappa,$ so $|\operatorname{rank}(x)|<\lambda <\kappa.$

Further, how does the regularity of $\kappa$ imply that $rank(x) < \kappa $?

I tried to make the proof a bit more explicit about how you can see that, but it's because $\operatorname{rank}(x)$ is the supremum of fewer then $\kappa$ ordinals less than $\kappa.$

The case of limit cardinal is much clearer (at least, I can see how the assummption that I stated above is used). The only unclear thing here is why there is $\kappa' < \kappa$ such that $hcard(x) < \kappa'$. The fact that $\kappa$ is a limit ordinal means that for any cardinal $\kappa_1 < \kappa$ there is a cardinal $\kappa_2$ with $\kappa_1 < \kappa_ 2 < \kappa$. But why does $hcard(x)$ have to do with $\kappa_2$ (or $\kappa_1$)?

If $\kappa$ is a limit cardinal, then for any set $A,$ if $|A|<\kappa$ then there's a (regular) cardinal $\lambda <\kappa$ with $|A|<\lambda.$ This is just what it means for $\kappa$ to be a limit cardinal... the cardinals below it are unbounded. (And we can take $\lambda$ to be regular since any successor is regular and the successor cardinals must also be unbounded below a limit cardinal.)

Is this general outline correct? Also, is there some kind of "base of induction" case that needs to be considered?

The infinite cardinals are a well-ordered class, so you can do induction on them like any other well-ordered class. You can do "strong induction" where you assume the property holds for all lesser elements and there is no base case, or you can "weak induction" where you do base case, successor case, and limit case. In the instance of infinite cardinals, $\aleph_0$ would be the base case.
