How to prove a cone is convex and closed? 
Let $A$ be a $m\times n$ matrix and consider the cones $G_0=\{d\in\mathbb R^n:Ad<0\}$ and $G'=\{d\in\mathbb R^n:Ad\le0\}$
Prove that $G'$ is a convex closed cone.

Lets see that $G'=\overline{ G'}.$ Note that this contention $G'\subset\overline{ G'}$ is always true. Let's see the other contention.
Suppose $d\in\overline{G'}$ and $d\not\in G'. $ Thus For every open ball $B$ with $d\in B$ we have that $B\cap G'\neq\emptyset$ and $ Ad>0$
How can I reach the contradition? Is this a good way to prove it or is there a better way? I don't know
And how to prove convexity?
Please help me please thanks

To prove $G'$ is closed from scratch without any advanced theorems. Following your suggestion, one way $G'\subset\overline{G'}$ is trivial, let's prove the opposite inclusion by contradiction. 
Let's start as you did by assuming that $\exists d\not\in G'$, $d\in\overline{G'}$. Since $d\not\in G'$, there exists one inequality among $Ad\le 0$ that is not true. In other words, there exists a row $a^T$ in the matrix $A$ such that $a^Td>0$. We denote $\rho=a^Td$ and show that for all $x$ that are close enough to $d$ we have $a^Tx>0$ too, which will mean that $x\not\in G'$ as well. Let's estimate
$$
a^Tx=a^Td+a^T(x-d)\ge\rho-|a^T(x-d)|\ge\rho-\|a\|\|x-d\|.
$$
From here we see that
$$
\|x-d\|<\frac{\rho}{\|a\|}=\epsilon\quad\Rightarrow\quad a^Tx>0.
$$
Conclusion: we have found an open ball $B$ around $d$
$$
B=\{x\colon \|x-d\|<\epsilon\}
$$
such that all $x\in B$ do not belong to $G'$, that is $B\cap G'=\emptyset$. This is a contradiction with the assumption that $d\in\overline{G'}$.
