Why is $e^{x}$ not uniformly continuous on $\mathbb{R}$?
If you are truly looking for a rigorous answer then you need to justify the "So, for $x$ large enough...".
For instance, here is a very rigorous solution along the lines you suggest:
Assume that $e^x$ is uniformly continuous on $\mathbb {R}$. Let $\epsilon = 1$. Thus there is $\delta >0$ such that for all $x,y\in \mathbb R$ if $|x-y|<\delta $ then $|e^x-e^y| < 1$. Let $a=\delta/2$. Since $\lim_{x\to\infty }e^x=\infty$ and since $e^a-1>0$ it follows that $\lim_{x\to \infty }e^x(e^a-1)=\infty$. Consequently, there is some $x\in \mathbb {R}$ such that $e^x(e^a-1)>1$. However, taking $y=x+a$ we have $|x-y|<\delta$ while $|e^x-e^y|=e^x(e^a-1)>1$, a contradiction.
Your proof looks good to me. There are only a few technicalities:
- You should consider an arbitrary $\delta>0$ because the contrapositive of $$\forall \epsilon>0,\ \exists\delta>0,\ \forall|x-y|<\delta,\ |f(x)-f(y)|<\epsilon$$ is $$\exists \epsilon>0,\ \forall\delta>0,\ \exists|x-y|<\delta,\ |f(x)-f(y)|\ge\epsilon.$$
- You should consider something like $e^{x+\delta/2}-e^x$ rather than $e^{x+\delta}-e^x$ because $|(x+\delta)-x|$ is not smaller than $\delta$.
- As Clayton pointed out in the comment section, if you claim that $(\star)$ does not hold when $x$ is large enough, you'd better explicitly write down an instance of $x$.