Showing that $\{\vec{x} : x_2 \geq e^{-x_1}\}$ is a convex set I am tasked to prove by the definition of convex sets that $S=\{\vec{x} : x_2 \geq e^{-x_1}\}$ is convex. It is simple to show this by considering the function $f(x) = e^{-x_2}-x_1$ and utilizing lower-level sets, but I would like to prove this directly by the definition of convex sets (i.e. the line segment between two points in $S$ is also in $S$).
So far, I've let $\vec{u}, \vec{v} \in S$, where $u=(u_1, u_2)$ and $v=(v_1, v_2)$ and established that our goal is to show that $\lambda u_2 + (1-\lambda)v_2 \geq e^{-(\lambda u_1 + (1-\lambda) v_1)}$.
From there, I've claimed that since $u_2 \geq e^{-u_1}$ and $v_2 \geq e^{-v_1}$, then $$\lambda u_2 + (1-\lambda)v_2 \geq \lambda e^{-u_1}+(1-\lambda)e^{-v_1}.$$
However, I don't know how to show that $\lambda e^{-u_1}+(1-\lambda)e^{-v_1} \geq e^{-(\lambda u_1 + (1-\lambda) v_2)}$ to conclude. So, how does one show that inequality?
 A: The statement of your problem and discussion in comments has confused me a bit.
As I understand your question, you want to prove your 2-dimensional subset $S$ is convex by verifying that for any two points in $S$ the line segment connecting them is contained in $S$.  Let $f:x\mapsto\exp(-x)$.  You have reduced your problem to showing that for all $u,v\in S$, and $\lambda\in[0,1]$ you have $$  \lambda f(u_1)+(1−\lambda)f(v_1)≥f(\lambda u_1+(1−\lambda)v_1).$$  This is precisely the defining  condition for $f$ to be  a convex function of a single variable.  So to complete your proof you need to show that $f$ is convex.
You seem to think that this argument is the same as showing that $S$ is a lower level set of the convex function $g(x,y)=\exp(-x)-y$ of two variables.  Granted, they are close, but in my opinion they are not identical. To me the "$f$" proof is simpler than the "$g$" proof because I cannot see how you can possibly show $g$ is convex without also showing (or knowing) that $f$ is convex.  The $f$ proof is, I think, more elegant, and contains more of the ideas about convexity you should keep in your head forever.
