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I am new to set theory and have been working through the proof that every measurable cardinal is Mahlo on page 135 of Jech's text. With the help of Asaf's comments (Measurable $rightarrow$ Mahlo), I have been able to make sense of the first half of the proof.

However, I found the second half (that argues by contradiction that $alpha < kappa : alpha text is regular in D$) quite terse, and cannot quite follow what is going on.

Could someone please provide a detailed version of Jech's proof or perhaps a detailed alternative proof (that mimics the proof that every measurable cardinal is inaccessible, which I believe I understand better).

Thank you in advance for your help.

Here is a proof leveraging ultrapowers. It generalizes well to all sorts of situations which is why I recommend learning it at some point:

In the following let $U$ be a normal ultrafilter on a measurable cardinal $kappa$ and let
$$
pi colon V to mathrmUlt(V;U)
$$
be the canonical ultrapower embedding (we regard $mathrmUlt(V;U)$ as transitive).

Claim. Let $X subseteq kappa$. Then $pi(X) cap kappa = X$.

Proof. For $xi < kappa$ we have
$$
xi in X iff pi(xi) = xi in pi(X).
$$
Q.E.D.

Claim. Let $C subseteq kappa$ be a club. Then $kappa in pi(C)$.

Proof. By elementarity
$$
mathrmUlt(V;U) models pi(C) text is a club in pi(kappa)
$$
and $pi(C) cap kappa = C$ is unbounded below $kappa < pi(kappa)$.

Thus
$$
mathrmUlt(V;U) models kappa in pi(C)
$$
and (by $Sigma_0$-absoluteness) hence $kappa in pi(C)$. Q.E.D.

Claim. $mathrmUlt(V;U) models kappa text is regular$.

Proof. $kappa$ is regular in $V$, $mathrmUlt(V;U) subseteq V$ and regularity is downward-absolute (a short cofinal sequence in $mathrmUlt(V;U)$ would also witness in $V$ that $kappa$ is singular). Q.E.D.

Now combine all of this:

Let $C subseteq kappa$ be a club. Then
$$
mathrmUlt(V;U) models kappa in pi(C) text and kappa text is regular .
$$
In particular
$$
mathrmUlt(V;U) models pi(C) text contains a regular cardinal.
$$
By the elementarity of $pi$ we obtain that
$$
V models C text contains a regular cardinal.
$$
Since $C$ was an arbitrary club in $kappa$, it follows that $kappa$ is Mahlo.

Here is a detailed explanation of Jech's proof.

Let $D$ be a normal measure on $kappa$. Suppose, towards a contradiction, that $kappa$ is not Mahlo. Then there is some club $C subseteq kappa$ such that
$$
C cap alpha < kappa mid mathrmcof(alpha) = alpha = emptyset.
$$

Since $D$ is normal, it contains all clubs. In particular $C in D$. Since $D$ is closed under intersections, we therefore must have that $ alpha < kappa mid mathrmcof(alpha) = alpha not in D$ and hence that
$$
alpha < kappa mid mathrmcof(alpha) < alpha = kappa setminus alpha < kappa mid mathrmcof(alpha) = alpha in D.
$$

By normality there is some $lambda < kappa$ such that

$$
E_lambda = alpha < kappa mid mathrmcof(alpha) = lambda in D.
$$

By replacing $E_lambda$ with $E_lambda setminus lambda$ we may and shall assume that $E_lambda cap lambda = emptyset$

For each $alpha in E_lambda$ fix a strictly increasing, cofinal function
$$
f_alpha colon lambda to alpha
$$
Now, for each $xi < lambda$, the function
$$
g_xi colon E_lambda to kappa, alpha mapsto f_alpha(xi)
$$
is decreasing. Hence there is some $A_xi in D$ and some $y_xi < kappa$ such that $f_alpha(xi) = y_xi$ for all $alpha in A_xi$.

Let

$$
A = bigcap_xi < lambda A_xi.
$$

Since $lambda < kappa$ we have that $A in D$.

Now let $alpha in A$. For all $xi < lambda$ we have $f_alpha(xi) = y_xi$ is independent of $alpha$ (by the construction of $A_xi$). But
$$
alpha = sup_xi < lambda f_alpha(xi) = y_xi
$$
is completely determined by the sequence $(y_xi mid xi < lambda)$.

Hence $A$ contains at most one element. This is a contradiction, since $D$ is non-principal.

It follows that $kappa$ is Mahlo after all!