I would like some feedback on my attempt / whether or not this is a valid approach as it differs from the other solutions I have been able to find. I am aware that similar questions have been asked about this question before, however, the solutions are different to mine and so I would like to know if this approach is valid.
Question
Prove $e^x$ is not uniformly continuous on $\mathbb{R}$
Attempt
Consider a proof by contradiction, where we assume that $e^x$ is uniformly continuous and therefore satisfies for all $\epsilon > 0$, there exists $\delta >0$ such that for real numbers $x$ and $y$:
$$|x-y|<\delta \implies |e^x-e^y|<\epsilon$$
We now consider $y=x+\frac{\delta}{2}$ which allows the LHS of the inequality to be satisfied.
Therefore, this should imply that:
$$|e^x-e^{x+\frac{\delta}{2}} |<\epsilon \iff|e^x(1-e^{\frac{\delta}2})|<\epsilon$$
This should hold for any real value of $x$. And so if we let $x=0$ then this inequality simplifies to:
$$|(1-e^{\frac{\delta}2})|<\epsilon$$
And so if we fix $\epsilon = 0.5|(1-e^{\frac{\delta}2})|$ then this contradicts the inequality above.
Therefore, we have shown that $e^x$ is not uniformly continuous by contradiction. $\Box$
I would be grateful for any feedback and guidance on ways in which I can improve the proof attempt.
The $\delta$ you choose depends on the $\varepsilon$ so you can't take $\varepsilon$ in terms of $\delta$.
However, if you fix $\varepsilon >0$ and suppose that there exists $\delta>0$ such that $$|x-y|<\delta \Rightarrow |e^x-e^y|<\varepsilon.$$ Take $y=x+\delta/2$ as you did, and now you can let $x$ tend to $+\infty$ to obtain a contradiction : $$\varepsilon\geq \lim_{x\rightarrow +\infty}e^x|1-e^{\delta/2}|=+\infty.$$