Attempt
First off, the cone of tangents of a set $S$ at a point $\overline{x}$ is the set of directions $d$ such that there exists a sequence of scalars $\{\lambda_k\}$ and vectors $\{x_k\}$ for $k=1,2,\dots,$ such that $\lambda_k>0$ for all $k$, $x_k\in{S}$ for all $k$, $x_k\to\overline{x}$ and $d=\lim_{k\to\infty}\lambda_k(x_k-\overline{x})$.
Now suppose $d\in T$ and $d\not\in G'$. As $d\not\in G',$ there exists $d$ such that $\nabla g_i(\overline x)^td>0$ which is equivalent to $-\nabla g_i(\overline x)^td\le0$. From here we deduce that $-\nabla g_i(\overline x)d=0,d\ge0$ has no solution by Gordan's Theorem.
Should I substitute the $d$ by the limit?
From here how can I reach the contradiction?
Could someone guide me please? Did I take a good way to prove it?
Thank you in advance.
It is easier here to do a direct proof without contradiction: let $d\in T$, we are to prove that $d\in G'$.
Start with Taylor's expansion at $\bar x$ for a $g_i(x)$, $i\in I$: $$ g_i(x_k)=g_i(\bar x)+\nabla g_i(\bar x)^T(x_k-\bar x)+o(|x_k-\bar x|). $$ We have
It becomes $$ \nabla g_i(\bar x)^T(x_k-\bar x)+o(|x_k-\bar x|)\le 0. $$ Multiply by $\lambda _k$ and let $k\to+\infty$. Since $\lambda_k(x_k-\bar x)\to d\in T$, the only think left for you to do here is to motivate that $$ \lambda_k\cdot o(|x_k-\bar x|)\to 0,\quad k\to+\infty $$ and conclude that $d\in G'$.