[proofplan] We use the stated function-and-embedding form of Woodin cardinal directly. A singular $\delta$ gives a cofinal sequence, and the Woodin closure condition produces a measurable critical point $\kappa<\delta$ whose cofinality is forced to be too small. For the strong-limit property, we choose the Woodin witness so that its critical point lies above an arbitrary $\alpha<\delta$; the standard external theorem that critical points of embeddings $j:V\to M$ are measurable, together with the standard external theorem that measurable cardinals are inaccessible, bounds $2^\alpha$ below $\delta$. [/proofplan] [step:Record the two external embedding facts used in the proof] We shall use the following two standard external theorems from large-cardinal theory. First, the critical-point measurability theorem says: if $M$ is a transitive proper class model of ZFC containing all ordinals and $j:V\to M$ is a nontrivial elementary embedding with critical point $\kappa$, then $\kappa$ is a measurable cardinal. In the form needed here, the normal measure is \begin{align*} U=\{X\subset\kappa: \kappa\in j(X)\}. \end{align*} The hypotheses required by this theorem are exactly transitivity of $M$, ZFC in $M$, containment of all ordinals in $M$, nontriviality of $j$, and existence of the critical point $\kappa$. Second, the measurable-inaccessibility theorem says: every measurable cardinal is an uncountable regular cardinal and a strong limit cardinal. Thus every measurable cardinal is an inaccessible cardinal. Every time these facts are invoked below, the Woodin witness supplies their hypotheses: the embedding is nontrivial, $M$ is transitive, $M$ is a proper class model of ZFC containing all ordinals, and the critical point is the stated ordinal $\kappa$. [/step] [step:Use the Woodin closure condition to rule out singular cofinality] Assume toward a contradiction that $\delta$ is singular. Let $\mu=\operatorname{cf}(\delta)$ denote the cofinality of $\delta$. Since $\delta$ is singular, $\mu<\delta$. Choose a strictly increasing cofinal map \begin{align*} c:\mu\to\delta. \end{align*} Define a function \begin{align*} g:\delta\to\delta \end{align*} by declaring $g(\xi)$ to be the least member of $c[\mu]$ strictly greater than $\xi$. This is well-defined because $c[\mu]$ is cofinal in $\delta$ and the ordinals are well-ordered. Define \begin{align*} f:\delta\to\delta \end{align*} by \begin{align*} f(\xi)=\max\{g(\xi),\mu+1\}. \end{align*} The value lies below $\delta$ because $g(\xi)<\delta$ and $\mu+1<\delta$. Apply the Woodin embedding property to $f$. We obtain $\kappa<\delta$, a transitive proper class model $M$ of ZFC containing all ordinals, and a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$, such that $f[\kappa]\subset\kappa$ and $V_{j(f)(\kappa)}\subset M$. Since the embedding is nontrivial, its critical point is nonzero. Hence $0<\kappa$, so $f(0)\in f[\kappa]$. The closure condition gives $f(0)<\kappa$, and because $f(0)\geq\mu+1$, we get \begin{align*} \mu<\kappa. \end{align*} By the critical-point measurability theorem, the hypotheses being supplied by the Woodin witness, $\kappa$ is measurable. By the measurable-inaccessibility theorem, $\kappa$ is regular. On the other hand, $f[\kappa]\subset\kappa$ implies $g[\kappa]\subset\kappa$, since $g(\xi)\leq f(\xi)$ for every $\xi<\delta$. For every $\xi<\kappa$, the ordinal $g(\xi)$ lies in $c[\mu]\cap\kappa$ and satisfies $\xi<g(\xi)$. Therefore $c[\mu]\cap\kappa$ is cofinal in $\kappa$. Since this cofinal set is contained in the image of the map $c:\mu\to\delta$, it has cardinality at most $\mu$, and hence \begin{align*} \operatorname{cf}(\kappa)\leq\mu. \end{align*} Together with $\mu<\kappa$, this contradicts the regularity of $\kappa$. Hence $\delta$ is regular. [guided] Suppose $\delta$ were singular. The goal is to turn a cofinal sequence in $\delta$ into a cofinal sequence in the critical point $\kappa$, contradicting the regularity of $\kappa$. Let $\mu=\operatorname{cf}(\delta)$. Since $\delta$ is singular, $\mu<\delta$. Choose a strictly increasing cofinal map \begin{align*} c:\mu\to\delta. \end{align*} For each $\xi<\delta$, define $g(\xi)$ to be the least element of $c[\mu]$ strictly above $\xi$. Such an element exists because $c[\mu]$ is cofinal in $\delta$, and it is least because ordinals are well-ordered. Thus we have a function \begin{align*} g:\delta\to\delta. \end{align*} Now define \begin{align*} f:\delta\to\delta \end{align*} by \begin{align*} f(\xi)=\max\{g(\xi),\mu+1\}. \end{align*} This function is designed so that any closure point $\kappa$ for $f$ must lie above $\mu$, while the values of $g$ below $\kappa$ still come from the original cofinal sequence $c$. Apply the Woodin embedding property to $f$. We get an ordinal $\kappa<\delta$, a transitive proper class model $M$ of ZFC containing all ordinals, and a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$, such that $f[\kappa]\subset\kappa$. Because $j$ is nontrivial, its critical point is nonzero, so $0<\kappa$. Therefore the value $f(0)$ belongs to $f[\kappa]$. The closure condition gives $f(0)<\kappa$, while the definition of $f$ gives $f(0)\geq\mu+1$. Hence \begin{align*} \mu<\kappa. \end{align*} The Woodin witness also supplies the hypotheses of the critical-point measurability theorem: $M$ is transitive, $M$ is a proper class model of ZFC containing all ordinals, and $j:V\to M$ is nontrivial with critical point $\kappa$. Hence $\kappa$ is measurable. By the standard theorem that measurable cardinals are inaccessible, $\kappa$ is regular. Now use the closure condition again. Since $g(\xi)\leq f(\xi)$ for every $\xi<\delta$ and $f[\kappa]\subset\kappa$, we have $g[\kappa]\subset\kappa$. For every $\xi<\kappa$, the value $g(\xi)$ belongs to $c[\mu]$ and satisfies $\xi<g(\xi)<\kappa$. Thus $c[\mu]\cap\kappa$ is cofinal in $\kappa$. This cofinal set is indexed by $\mu$, so \begin{align*} \operatorname{cf}(\kappa)\leq\mu. \end{align*} But $\mu<\kappa$, so this contradicts regularity of $\kappa$. Therefore $\delta$ cannot be singular, and $\delta$ is regular. [/guided] [/step] [step:Show that $\delta$ is uncountable and has room above every smaller ordinal] Define the constant zero map \begin{align*} z:\delta\to\delta \end{align*} by $z(\xi)=0$. Applying the Woodin embedding property to $z$ gives a nontrivial elementary embedding $j_0:V\to M_0$ with critical point $\kappa_0<\delta$, where $M_0$ is a transitive proper class model of ZFC containing all ordinals. By the critical-point measurability theorem, $\kappa_0$ is measurable. By the measurable-inaccessibility theorem, $\kappa_0$ is uncountable. Since $\kappa_0<\delta$, the cardinal $\delta$ is uncountable. Because $\delta$ is now known to be an uncountable regular cardinal, it is a limit cardinal and, as an ordinal, has no immediate predecessor. Therefore for every $\alpha<\delta$, \begin{align*} \alpha+2<\delta. \end{align*} [/step] [step:Force a Woodin witness whose critical point lies above a chosen ordinal] Fix an ordinal $\alpha<\delta$. By the previous step, $\alpha+2<\delta$. Define a function \begin{align*} f_\alpha:\delta\to\delta \end{align*} by \begin{align*} f_\alpha(\xi)=\alpha+2 \end{align*} when $\xi\leq\alpha+2$, and by \begin{align*} f_\alpha(\xi)=0 \end{align*} when $\alpha+2<\xi<\delta$. This is a well-defined function into $\delta$ because both $0$ and $\alpha+2$ are below $\delta$. Apply the Woodin embedding property to $f_\alpha$. We obtain $\kappa<\delta$, a transitive proper class model $M$ of ZFC containing all ordinals, and a nontrivial elementary embedding $j:V\to M$ with critical point $\kappa$, such that $f_\alpha[\kappa]\subset\kappa$ and $V_{j(f_\alpha)(\kappa)}\subset M$. Since $j$ is nontrivial, $0<\kappa$. Hence $0<\kappa$ and $0\leq\alpha+2$ imply $f_\alpha(0)=\alpha+2$ belongs to $f_\alpha[\kappa]$. The closure condition gives \begin{align*} \alpha+2<\kappa. \end{align*} Thus \begin{align*} \alpha<\kappa<\delta. \end{align*} [guided] Fix $\alpha<\delta$. We want a Woodin witness whose critical point is above $\alpha$, because measurable critical points have strong-limit cardinal arithmetic below them. The technical point is that the constant value used to force the closure point must itself be below $\delta$. The previous step established that $\delta$ is an uncountable regular cardinal, so $\delta$ is a limit ordinal and \begin{align*} \alpha+2<\delta. \end{align*} Define \begin{align*} f_\alpha:\delta\to\delta \end{align*} by setting $f_\alpha(\xi)=\alpha+2$ for $\xi\leq\alpha+2$, and $f_\alpha(\xi)=0$ for $\alpha+2<\xi<\delta$. This declaration gives a genuine function into $\delta$ because both possible values, $0$ and $\alpha+2$, are ordinals below $\delta$. Apply the Woodin embedding property to this function. We obtain an ordinal $\kappa<\delta$ and a nontrivial elementary embedding $j:V\to M$ into a transitive proper class model of ZFC containing all ordinals, with critical point $\kappa$, such that $f_\alpha[\kappa]\subset\kappa$. Since the embedding is nontrivial, its critical point is nonzero. Hence $0<\kappa$, so $0$ lies in the initial segment whose image is $f_\alpha[\kappa]$. By the definition of $f_\alpha$, \begin{align*} f_\alpha(0)=\alpha+2. \end{align*} The closure condition $f_\alpha[\kappa]\subset\kappa$ then gives \begin{align*} \alpha+2<\kappa. \end{align*} Combining this with the witness condition $\kappa<\delta$ yields \begin{align*} \alpha<\kappa<\delta. \end{align*} [/guided] [/step] [step:Bound each power set below $\delta$] Keep the ordinal $\alpha<\delta$ and the witness $j:V\to M$ with critical point $\kappa$ from the previous step. The Woodin witness supplies the hypotheses of the critical-point measurability theorem, so $\kappa$ is measurable. By the measurable-inaccessibility theorem, $\kappa$ is a strong limit cardinal. Since $\alpha<\kappa$, the definition of strong limit cardinal gives \begin{align*} 2^\alpha<\kappa. \end{align*} The witness condition gives $\kappa<\delta$, and therefore \begin{align*} 2^\alpha<\delta. \end{align*} Because $\alpha<\delta$ was arbitrary, $\delta$ is a strong limit cardinal. [/step] [step:Combine the three cardinal properties] We have proved that $\delta$ is uncountable, regular, and a strong limit cardinal. These are exactly the defining properties of an inaccessible cardinal. Hence $\delta$ is inaccessible. [/step]