Given continuous $u:\Omega\to\mathbb{R}$ and $x\in \Omega$, define $$J^{1,+}u(x)=\{p\in\mathbb{R}^n\mid u(x+z)\le u(x)+(p,z)+o(|z|)\},$$ as $z\to 0$. Here $f(z)\le o(|z|)$ means there exists a continuous $\sigma:[0,\infty)\to [0,\infty)$ such that $\sigma(0)=0$ and $$ \frac{f(z)}{|z|}\le\sigma(|z|),\quad |z|>0.$$ I want to show $J^{1,+}u(x)$ is closed.
Let $p_n\in J^{1,+}u(x)$ such that $p_n\to p$. Then by definition, I need to find continuous $\sigma:[0,\infty)\to [0,\infty)$ such that $$ \frac{u(x+z)- u(x)-(p,z)}{|z|}\le \sigma(|z|).\tag{1} $$
Below is what I have tried:
By assumption, there exists $\sigma_n$ such that for all $n$ $$ \frac{u(x+z)- u(x)-(p_n,z)}{|z|}\le\sigma_n(|z|),\quad |z|>0. $$ which implies that $$ \frac{u(x+z)- u(x)-(p,z)}{|z|}\le |p-p_n|+\sigma_n(|z|). $$ Using the above inequality, I can show $$ \limsup_{z\to 0}\frac{u(x+z)- u(x)-(p,z)}{|z|}\le 0. \tag{2} $$ But the problem is I don't know how to find a fixed continuous $\sigma$ for $p$ satisfying the requirement (1). Because for $|z|<r$, I can choose $|p_n-p|<r$, but the $\sigma_n$ also changes with $r$.
You basically have it. You need to show that for every $\varepsilon>0$ there exists $\delta>0$ such that
$$\frac{u(x+z) - u(x) - (p,z)}{|z|} < \varepsilon$$
whenever $|z|< \delta$. Just choose $n$ large enough so that $|p-p_n| < \varepsilon/2$ and then choose $\delta>0$ so that $\sigma_n(|z|)<\varepsilon/2$ for $|z|< \delta$.