First, what basically distinguishes the definitions of convex, affine and cone, is the domain of the coefficients and the constraints that relate them.
Let us starts by the first part:
any subspace is affine, which means, if we have:
$x_1, x_2 \in V$, where $V$ is a subspace; therefore any linear combination of these two vectors must lie in $V$. That is, if we have two coefficients $\theta_1, \theta_2 \in \mathcal{R}$, then, $\theta_1x_1 + \theta_2x_2 \in V$.
The definition of affine sets tells us if $x_1,x_2$ are in an affine set, their linear combination must also lie in the same set, with the condition the coefficients must sum to 1, that is $\theta_1 + \theta_2 = 1$. Now, assume we have chosen $\theta_2 = 1- \theta_1$, therefore the combination $\theta_1x_1 + (1 - \theta_1)x_2 \in V$
Therefore any subspace is affine, since we have the freedom to choose the coefficients to sum to 1.
Now why a subspace is a convex cone.
Notice that, if we choose the coeficientes $\theta_1, \theta_2 \in \mathcal{R}_+$, we actually define a cone, and if the coefficients sum to 1, it is convex, therefore it is a convex cone.
I found my error. Just needed to take a break and return to the problem. The dual cone of the cone $K$ I used in my example is in fact not equal to $R_{+}^{2}$. What finally made me realize this is that the dual cone of $R_{+}^{2}$ is itself. But I know that calculating the dual cone twice is supposed to return you the original cone.
I was making a silly error. The dual cone of the 45 deg cone centered about the $x = y$ ray for $x > 0$ is actually the 135 deg cone centered about the same ray.
Here is the corrected drawing. Any vector along the green vector will clearly result in $\lambda^{T}x \gt \lambda^{T}y$.
Corrected Example
Best Answer
Let $y\in V^*$ and assume that there exists $v\in V$ such that $y^Tv >0.$ Since $V$ is a subspace, it follows that $-v \in V.$ But then
$$0\leq y^T(-v)= -y^Tv<0,$$ a contradiction.