Is a convex salient cone necessarily contained in an open half-space? A cone $C$ in $\Bbb R^n$ is said to be salient if it does not contain any pair of opposite nonzero vectors; that is, if and only if $C \cap (-C) \subset \{0\}$.
Obviously, a cone $C$ such that that $C\setminus\{0\}$ is contained in an open half-plane is salient.
I suspect the converse might be true for convex cones but I have found no reference to this result.

On the other hand, this sentence from Wikipedia seems to make a difference between the two notions, depending on how you interpret the or.

The term proper cone is variously defined, depending on the context. It often means a salient and convex cone, or a cone that is contained in an open halfspace of V.

 A: As a matter of fact even more is true, at least when $\overline C$ is salient (in particular if $C$ is closed and salient):
If $C$ is a convex cone in $\mathbb R^n$ such that $\overline C$ is salient and $C \neq \{0\}$, then there are coordinates $(x_1, \ldots, x_n)$ in $\mathbb R^n$ such that $C \setminus \{0\}$ is contained in a positive orthant, that is in the subset
$$\{(x_1, \ldots, x_n) \in \mathbb R^n: x_i > 0, 1 \le i \le n \}.$$
Here is a proof. Recall that the dual cone of $C$ is 
$$C^* = \{x \in \mathbb R^n : <x, y> \ge 0 \ \forall y \in C\}.$$ 
First let us see that $C^*$ is full-dimensional. In fact, if not, $C^*$ is contained in a hyperplane $H$ and, as $C^*$ contains $0$, it follows that there exists a $v \neq 0$ such that $H = v^{\perp} = \{x \in \mathbb R^n : <x, v> = 0\}$. Now for every $x \in C^*$ we have that $x \in H$, whence $<x, v> = 0$. Therefore also $<x, -v> = 0$. We deduce that both $v$ and $-v$ belong to $(C^*)^*$. As is well-known (see for example, Rockafellar, Convex Analysis, page 121), $(C^*)^* = \overline C$, so $\overline C$ is not salient. 
Hence $C^*$ is full-dimensional and nonempty, therefore, by Thm.6.2 (Rockafellar), the interior of $C^*$ is nonempty. Hence we can find a basis $\{v_1, \ldots, v_n \}$ such that each $v_i$ lies in the interior of $C^*$, that is $<v_i, y> > 0 \ \forall y \in C \setminus \{0\}$. Now pick a basis $\{e_1, \ldots, e_n \}$ of ${\mathbb R}^n$ dual to the basis $\{v_1, \ldots, v_n \}$, that is such that $<v_i, e_j> = \delta_{ij}$ for all $i, j$. Then, for any $y \in C \setminus \{0\}$ we can write $y = x_1 e_1 + \ldots + x_n e_n$ with
$0 < <v_i, y> = x_i$ for all $i$.
A: It follows from the separating hyperplane theorem that any convex proper subset of $\mathbb R^n$ is contained in an open half space.  So, this holds true for convex cones in particular, even if they aren't salient (as long as the cone is a proper subset of $\mathbb R^n$).
A: Let us make some remarks, also in response @littleO.
First the notation: we let $C \subseteq  \mathbb R^n$ be a cone such that, to avoid trivialities, $C \neq \emptyset, \{0\}$.
Also, given $v \in \mathbb R^n \setminus \{0\}$ and $c \in \mathbb R$ we set
$$H_{v,c} = \{x \in \mathbb R^n : <x,v> = c \}, H_{v,c}^+ = \{x \in \mathbb R^n : <x,v> > c \}, H_{v,c}^- = \{x \in \mathbb R^n : <x,v> < c \},$$ 
$$H_v = H_{v,0}, H_v^+ = H_{v,0}^+, H_v^- = H_{v,0}^-.$$
Remarks:
(1) If $C \setminus \{0\} \subseteq H_v^+$, then $C$ is salient.
In fact, for any $x \in  C \setminus \{0\}$ we have $<x,v> > 0$, whence $<-x,v> < 0$, so that $-x \not\in C$, that is $C$ is salient.
(2) If $c \neq 0$, $C \setminus \{0\} \subseteq H_{v,c}^+$, does not, in general, imply that $C$ is salient.
In fact, take the closed half-plane $C = H_{e_2} \cup H_{e_2}^+$ in $\mathbb R^2$. Then $C$ is a closed convex nonsalient cone, but clearly $C \subseteq H_{e_2,-1}^+$.
(3) Even if $C$ is convex, the converse asked by Rhubarbe (or the converse of (1)) is, in general, false. That is, in general, if $C$ is a convex salient cone, it does not follow that there exists $v \in \mathbb R^n \setminus \{0\}$ such that $C \setminus \{0\} \subseteq H_v^+$.
In fact, let  $n=2$ and let
$$C = H_{e_2}^+ \cup \{(x_1,x_2) \in \mathbb R^2 : x_1 \ge 0, x_2=0\}.$$ 
Then $C$ is clearly a convex salient cone, but if there were a  $v = (v_1,v_2)\in \mathbb R^2 \setminus \{0\}$ such that $C \setminus \{0\} \subseteq H_v^+$, then, since $(1,0) \in C \setminus \{0\}$ we have $v_1 > 0$. But also $(-1,t) \in C \setminus \{0\}$ for every $t > 0$, whence $-v_1+tv_2 = <(-1,t),v> > 0$, so that $-v_1 \le 0$, a contradiction.
(4) The problem in (3) is that $\overline C$ is not salient. As we have seen in my previous answer, when $\overline C$ is salient, we actually have that $C \setminus \{0\}$ is contained in a positive orthant.
On the other hand, as remark (3) suggests, one should be able to remove a face from $C$ and get the converse asked by Rhubarbe. 
I give a proof below, using, as suggested by littleO, the hyperplane separation theorem.
(5) Let $C$ be a cone. If there exists a hyperplane $H = H_v$ such that $C \cap H$ is salient and $C \setminus C \cap H \subseteq H^+$, then $C$ is salient. If $C$ is convex and salient, then the converse holds.
Assume first that $H$ as above exists. Let $x \in C$ be such that $-x \in C$. If $x \not\in H$ then $x \in C \setminus C \cap H$, so that
$x \in H^+$, whence $<x,v> > 0$. But then $<-x,v> < 0$, and then also $-x \in C \setminus C \cap H$, therefore $-x \in H^+$, that is $<-x,v> > 0$, a contradiction. Therefore $x \in H$, and then $<x,v>=0$, whence also $<-x,v>=0$, that is $-x \in H$. Hence $x, -x \in C \cap H$, so that $x = 0$ and we have proved that $C$ is salient.
Finally assume that $C$ is convex and salient. First note that $C \setminus \{0\}$ is also convex: if $x, y \in C \setminus \{0\}$ and $t \in [0,1]$, then, as $C$ is convex, $tx+(1-t)y \in C$. If $tx+(1-t)y \not\in C \setminus \{0\}$, then $tx+(1-t)y = 0$, so that $t \neq 0, 1$ and 
$$\frac{t-1}{t}y = x \in C.$$
Since $\frac{1-t}{t} > 0$, we also have that 
$$-x = \frac{1-t}{t}y \in C,$$
a contradiction. Therefore $C \setminus \{0\}$ is convex and disjoint from $\{0\}$. By the hyperplane separation theorem there are a nonzero $v \in \mathbb R^n$ and some $c \in \mathbb R$ such that
$$C \setminus \{0\} \subseteq H_{v,c} \cup H_{v,c}^+ \ \mbox{and} \ \{0\} \subseteq H_{v,c} \cup H_{v,c}^-.$$
Now the second condition gives $0 = <0,v> \le c$. On the other hand the first implies, by taking closure, that $0 \in \{x \in \mathbb R^n : <x,v> \ge c \}$, that is $0 = <0,v> \ge c$. Therefore $c = 0$ and setting $H = H_v$ we have that $C \setminus \{0\} \subseteq H \cup H^+$, whence $C \setminus C \cap H \subseteq H^+$ and, of course, $C \cap H$ is salient. 
