My reasoning is, the vector space being a 'flat plane', a straight line joining any two points on it will always be contained in $\mathbb{R}^2$. And it has no points in it which cannot be on the straight line connecting two other points in the set (since the plane is infinite)
I am new to convex analysis and optimization techniques, so my line of thinking is very informal. Please let me know whether or not my approach is correct.
On
You're right that it is convex. Taking any $\vec x,\vec y\in\Bbb R^2,$ we have for all $\alpha\in(0,1)$ that $\alpha\vec x,(1-\alpha)\vec y\in\Bbb R^2$ by closure under scalar multiplication, and so $\alpha\vec x+(1-\alpha)\vec y\in\Bbb R^2$ by closure under addition.
As for extreme points, note that if $\vec x\ne\vec 0,$ then $\vec x$ lies on the open line segment between $\frac12\vec x$ and $\frac32\vec x,$ so cannot be extreme. I will leave it to you to show (by example) that $\vec 0$ cannot be an extreme point, either.
For any point $E \in \mathbb{R}^2$, because $\mathbb{R}^2$ is closed under addition and scalar multiplication, there exist two points (vectors) $x,y \in \mathbb{R}^2$ and an $\alpha \in (0,1)$ such that $E = \alpha x + (1-\alpha)y \in \mathbb{R}^2$. Thus, the space does not contain any extreme points.