(1) If $X$ is a some subset in $\mathbb{R}^n$,
$N=1$ : For $x\in X$, $1x\in {\rm conv}\ X$ so that $$ X\subset {\rm conv}\ X $$
$N=2$ : For $x,\ y\in X$, $$ \lambda_1x+\lambda_2 y \in {\rm conv}\ X,\ \lambda_1+ \lambda_2=1,\ \lambda_i\geq 0 $$ That is any line between points in $X$ is in ${\rm conv}\ X$.
(2) And $Y:={\rm conv}\ X$ is convex : If $x,\ y\in Y$, then $$ x= \sum_{i=1}^k\lambda_i x_i,\ y=\sum_{j=1}^{l} \tau_j y_i,\ x_i,\ y_j\in X $$
Then for $a,\ b\geq 0,\ a+b=1$, then we have $$ ax+by = \sum_{i=1}^ka\lambda_i x_i + \sum_{j=1}^{l} b\tau_j y_i $$ And note that $$
\sum_{i=1}^ka\lambda_i + \sum_{j=1}^{l} b \tau_j =1 $$
(3) $Y$ is smallest set which is a convex set containing $X$ : If $Z$ is a convex set containing $X$, then any line between points in $X$ is in $Z$. And if $$\lambda_i\geq 0,\ \sum_{i=1}^3\lambda_i=1,\ x_i\in X $$ then $$ x:=\sum_{i=1}^3\lambda_i x_i=
\lambda_1 x_1 + (1-\lambda_1) \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } x_i $$ Note that $$ \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } =1 $$
So $ \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } x_i \in Z$ so that $x\in Z$ Continuously we have $ \sum_{i=1}^k\lambda_i x_i\in Z$ so that ${\rm conv}\ X\subset Z$.
(4) By definition $$ {\rm conv}\ X\subset {\rm cone}\ X $$
Let $$ Z:= \{ x\in \mathbb{R}^n\mid tx \in {\rm conv}\ X,\ {\rm
some}\ t\geq 0\} \cup \{0\}$$ Then
$$ Z = {\rm cone}\ X$$
Proof : $Z\subset {\rm cone}\ X $ : $$ x\in Z\rightarrow tx \in {\rm
conv}\ X \rightarrow tx=\sum_i c_i x_i,\ \sum_i c_i =1
$$
so that $$ x= \sum_i \frac{c_i}{t} x_i \in {\rm cone}\ X $$
${\rm cone}\ X\subset Z $ : Let $x=\sum_i c_i x_i \in {\rm cone}\
X,\ c:=\sum_i c_i$. Then $$ \frac{x}{c} = \sum_i \frac{ c_i}{c} x_i
\in {\rm conv}\ X $$ Let $t= \frac{1}{c}$ so that $x\in Z $.
Nothing of this sort is true for most standard cones (e.g. the natural cone in classical function spaces). Consider the following examples.
Example 1. Let $X = C_{\mathbb{R}}[0,1]$ with its usual cone. Then an extreme ray corresponds with a function $f \in X_+ \setminus \{0\}$ such that $0 \leq g \leq f$ implies $g = \alpha f$. But any non-zero function must be non-zero on some open interval, so for any $f \gneq 0$ there is a plethora of functions lying between $0$ and $f$. Conclusion: there are no extreme rays.
Example 2. Let $Y = \ell_{\mathbb{R}}^\infty$ with its usual cone. The extreme rays are the standard basis vectors $e_i$, so the closed convex cone they generate is only $(c_0)_+$.
Example 3. Similarly, let $Z = B(\ell_{\mathbb{C}}^2)^{\text{sa}}$ (the self-adjoint operators $\ell_{\mathbb{C}}^2 \to \ell_{\mathbb{C}}^2$) with the positive semidefinite cone. The extreme rays are the rank one orthogonal projections. The closed cone they generate is $K(\ell_{\mathbb{C}}^2)_+$, the cone of compact positive semidefinite operators.
Similarly, the statement fails for many spaces of differentiable or Lebesgue integrable functions on some domain $U \subseteq \mathbb{R}^n$. (It is however true for most sequence spaces, for instance $\ell_{\mathbb{R}}^p$ with $1 \leq p < \infty$.)
To get sufficient criteria for the statement to be true, I guess one must resort to Krein–Milman type theorems (e.g. assume that the cone has a weakly compact base). For more on this, see §3.8 and §3.12 in [Jam70]. (Warning before reading Jameson's book: in the ordered vector spaces community, cone means proper/pointed convex cone — see §1.1.) In particular:
Proposition. Let $E$ be a locally convex space, and let $E_+ \subseteq E$ be a convex cone. If $E$ has an interior point, then the (topological) dual cone $E_+' \subseteq E'$ is the weak-$*$ closed convex cone generated by its extreme rays.
Proof. Combine Theorem 3.8.6, Theorem 3.12.8 and Corollary 3.12.9 from [Jam70]. $\hspace{18mm}\Box$
References.
[Jam70]: Graham Jameson, Ordered Linear Spaces, Springer Lecture Notes in Mathematics 141, 1970.
Best Answer
Let $C^\ominus = \{ u\in E^* \,|\, \langle u,C\rangle \leq 0\}$ and let $H = \{u\in E^* \,|\, \langle u,-y\rangle \leq 0\}$.
If the interior of $C^\ominus$ meets $H$, or if $C^\ominus$ meets the interior of $H$, then we have the sum rule $\partial (\iota_{C^\ominus}+\iota_{H})= \partial \iota_{C^\ominus} + \partial \iota_{H}$. In that case, evaluating at $0$, gives
$$ (C^\ominus\cap H)^\ominus = C+H^\ominus = C+\mathbb{R}_+y\;\;\text{is closed.}$$
Another case arises when $C$ is a polyhedral cone. All the above is true in $\mathbb{R}^n$ and in Hilbert space, and likely in Banach space. I don't know about your very general setting.