Is it possible to prove $|V_\kappa|=\kappa$ for strongly inaccessible $\kappa$ without AC? First, let me note that my definition of "strongly inaccessible" is that a nonzero ordinal $\kappa$ is strongly inaccessible if ${\rm cf}(\kappa)=\kappa$ and $\forall\alpha<\kappa, {\cal P}(\alpha)\prec\kappa$ in the sense of an injection from ${\cal P}(\alpha)$ to $\kappa$. (Thus $\omega$ is strongly inaccessible, although it is probably not important.)
The only way I know to prove that $|V_\kappa|=\kappa$ is to prove that $V_\alpha\prec\kappa$ for all $\alpha<\kappa$ by induction. If $V_\beta\prec\kappa$, then $|V_\beta|=\gamma$ for some $\gamma<\kappa$, and so $|V_{\beta+1}|=|{\cal P}(V_\beta)|=|{\cal P}(\gamma)|<\kappa$ since $\kappa$ is a strong limit. For limit ordinals $\alpha<\kappa$, $V_\alpha=\bigcup_{\beta<\alpha}V_\beta$, so since $V_\beta\prec\kappa$, $V_\alpha\preceq\alpha\times\kappa\prec\kappa\times\kappa\approx\kappa$ (where the inequalities correspond to injections and $\alpha\times\kappa$ is a cross product) by AC. Then we can finish by noting again that $V_\alpha\prec\kappa$ implies $V_\kappa=\bigcup_{\alpha<\kappa}V_\alpha\preceq\bigsqcup_{\alpha<\kappa}\kappa=\kappa\times\kappa\approx\kappa$, and $\kappa\subseteq V_\kappa$ so that $V_\kappa\approx\kappa$ by Schroder-Bernstein.
In the proof above, I used AC twice, in the claim that $\forall \alpha<\kappa,A_\alpha\preceq B$ implies $\bigcup_{\alpha<\kappa}A_\alpha\preceq\kappa\times B$. Is it possible to generalize the approach of this question to avoid AC altogether? Also, it appears I only used the strong limit property above, without ever using the fact that $\kappa$ is regular. That can't be right; is it true that $|V(\beth_\omega)|=\beth_\omega$?
Edit: The above transfinite induction proof has a flaw in it in the limit step. For limit ordinals $\alpha<\kappa$, $V_\alpha=\bigcup_{\beta<\alpha}V_\beta$, so since $V_\beta\prec\kappa$, $V_\alpha\preceq\alpha\times\kappa\preceq\kappa\times\kappa\approx\kappa$ by AC, but the inequality is not strict. To make it strict, we note that if $|V_\alpha|=\kappa$, then the function $f:\alpha\to\kappa$ defined by $f(\beta)=|V_\beta|$ is a cofinal map, so ${\rm cf}(\kappa)=\kappa\le\alpha$, a contradiction. Thus $|V_\alpha|<\kappa$. So it is not true that strong limits are sufficient to ensure that $|V_\kappa|=\kappa$.
 A: First of all, assuming choice one can show that $|V_{\omega+\alpha}|=\beth_\alpha$ (this addition is ordinal addition!), so we have $|V_\alpha|=\alpha$ if and only if $|\omega+\alpha|=|\alpha|=\beth_\alpha$ (note that this implies $\omega+\alpha=\alpha$ as ordinals). Inaccessible cardinals are indeed $\beth$ fixed points, but $\beth_\omega$ is certainly not. 
So while there are fixed points which are singular, you still can't do that for any strong limit cardinal.

Now, to your main question. You can't adapt the proof from $\omega$ to an uncountable inaccessible $\kappa$, because that would mean that $V_\kappa\models\sf ZF+\text{Global Choice}$, because we can find a well-ordering of $V_\kappa$ internally definable within $V_\kappa$.
If we assume that $V_{\kappa+1}$ can be well-ordered, then this is not very difficult. The proof follows from Theorem 11 and Proposition 4 from the paper:

Blass, Andreas, Ioanna Dimitriou, and Benedikt Lowe. Inaccessible cardinals without the axiom of choice. Fundamenta mathematicae 194.2 (2007): 179-189.

You have to slightly adjust your construction as presented there, but it works. Alternatively, you can use the same proof as Jech gives (and attributes to Rubin & Rubin) in Theorem 9.1 (b) in his Axiom of Choice book.

Now, I haven't sat down to verify all the details, but it should probably work as a counterexample. And a contrived one too. The idea is based on Joel David Hamkins proof that $\sf ZFC$ does not prove the universe can be linearly ordered.
Let $V$ be a model of $\sf ZFC$ and $\kappa\in V$ is inaccessible (for simplicity, assume $\sf GCH$ holds below $\kappa$). We consider the forcing $\Bbb P$ which add two subsets to each power set of a regular $\lambda<\kappa$ with functions whose domains are of cardinality $<\lambda$. 
So a condition in $\Bbb P$ is a function $p\colon\kappa\times2\times\kappa\to 2$ with the following properties:


*

*If $(\alpha,i,\beta)\in\operatorname{dom}(p)$ then $\beta<\aleph_{\alpha+1}$, and $i<2$.

*The set $\{\beta<\aleph_{\alpha+1}\mid (\alpha,i,\beta)\in\operatorname{dom}(p)\}$ has size of at most $\aleph_\alpha$.

*$|p|<\kappa$.


Now consider the automorphisms which "switch" the middle coordinate, $i$, separately between each $\alpha$. Informally, if $A_\alpha$ and $B_\alpha$ are the new subsets of $2^{\aleph_{\alpha+1}}$ then we just switch these two.
And let $\cal F$ be the filter of subgroups of bounded permutations, that is things which are fixed by a permutations which leave the $i$ coordinate intact for $i>i_0$ for some $i_0$. And let $N$ be the symmetric extension induced by $\cal F$.
It is not hard to show that $\kappa$ is still inaccessible in $N$. And it is not hard to show that the set $R=\{A_\alpha,B_\alpha\mid\alpha<\kappa\}$ is also in $N$. But it cannot be linearly ordered in $N$. 
Since $R\subseteq V_\kappa$ we have that it is impossible that $|V_\kappa|=\kappa$ in $N$.
