I try this proof by contradiction that
Assuming $f:\mathbb{R} \rightarrow \mathbb{R}$, $f(x)=e^{x^2}$ is uniformly continuous.
By definition we have $\forall \varepsilon>0$ $\exists \delta>0$ $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$.
Choosing $\varepsilon=1$,we have $|x-y|<\delta \Rightarrow |f(x)-f(y)|<1$.
If $x<y$, than $|f(x)-f(y)|=|e^{y^2}-e^{x^2}|<|e^{(x+\delta)^2}-e^{x^2}|=|e^{x^2}(e^{\delta^2+2x\delta}-1)|<1$.
But for $x \rightarrow +\infty$, we have $|e^{x^2}(e^{\delta^2+2x\delta}-1)| \rightarrow +\infty$, which is a contradiction.
For $x>y$ it is the same.
May I ask if this is reasoning correct and sufficient?
Many thanks!