This is basically a proof verification question. I want to find a model $\mathrm{ZFC+V\neq HOD}$ where $\mathrm{HOD}$ is the class of heriditarilly ordinal definable sets. The idea is to add one Cohen real and show that it is not ordinal definable in the extension. So let $M$ be a ground model and $P\in M$ be Cohen forcing. Let $G$ be $M$-generic for $P$ and let $c=\bigcup G$.
Assume for a contradiction that $c\in\mathrm{OD}$. Then since there is a well-order $<_{\mathrm{OD}}$ of $\mathrm{OD}$ definable without parameters, we can define $c$ as the $\alpha$-th element of $<_{\mathrm{OD}}$ for some ordinal $\alpha\in M$. In particular there is an $\in$-formula $\varphi(x,y)$ with (this is the part which I'm not sure about)
\begin{equation}\tag{1} M[G]\models\forall i<\omega(c(i)=1 \leftrightarrow \varphi(i,\alpha)) \end{equation}
By the forcing theorem we can take $p\in G$ with
\begin{equation}\tag{2} p\Vdash\forall i<\check\omega(\bigcup \dot G(i)=1 \leftrightarrow \varphi(i,\check\alpha)) \end{equation}
where $\dot G$ is the canonical name for the generic filter. Now take any $n\notin\operatorname{dom}(p)$ and consider the embedding $\pi_n:P\to P$ with $$\pi_n(q):\operatorname{dom}(q)\to2 \qquad\text{with}\qquad \pi_n(q)(i)=\begin{cases}q(i) & i\neq n\\ 1-q(i) & i=n\end{cases}$$
Let $G':=\pi_n[G]$ and $c':=\bigcup G'$. Then (3) $G'$ is $M$-generic for $P$ (because $\pi_n$ is an embedding) with $p\in G'$ and (4) $M[G']=M[G]$ because the filters are easily definable from one another. Therefore \begin{align*} c(n)=1 &\iff M[G]\models\varphi(n,\alpha) \tag{by (1)}\\ &\iff M[G']\models\varphi(n,\alpha) \tag{by (4)}\\ &\iff c'(n)=1 \tag{by (2) and (3)} \end{align*} a contradiction.
Is this proof correct? Can some steps be done better/more elegantly? Are there other nice models of $\mathrm{V\neq HOD}$?
Yes, the proof seems fine.
You can also check out Why does $\mathsf{HOD}^{V[G]} \subseteq \mathsf{HOD}^V$ hold for weakly homogeneous forcings? where it is shown that if $\Bbb P$ is a weakly homogeneous forcing (e.g. the Cohen forcing, various Levy collapses, and many other natural forcings), then $\rm HOD^{\it V[G]}\subseteq HOD^{\it V}(\Bbb P)$.
In particular, if $V=\rm HOD$ holds in the ground model, then the result follows. So practically any homogeneous forcing over $L$ will violate $V=\rm HOD$, for example.