I am self-learning Real Analysis from Stephen Abbott's, Understanding Analysis. I'd like someone to verify, if my work for the below exercise problem checks out, whether the proof is correct.
[Abbott 6.2.7] Let $f$ be uniformly continuous on $\mathbf{R}$ and define a sequence of functions by $f_n(x) = f\left(x + \frac{1}{n}\right)$. Show that $(f_n)$ converges to $f$ uniformly. Give an example to show that this proposition fails if $f$ is only assumed to be continuous and not uniformly continuous.
Pick an arbitrary $\epsilon > 0$.
Since $f$ is uniformly continuous on $\mathbf{R}$, $(\exists \delta(\epsilon) > 0)$, such that $\forall x,y \in \mathbf{R}$ satisfying $|x - y| < \delta$, it follows that $|f(x) - f(y)| < \epsilon$.
Let $N = \frac{1}{\delta}$. Consider all $x,y \in \mathbf{R}$, where $y = x + \frac{1}{n}; n \in \mathbf{N}$ satisfying $|y - x| < \delta$, that is $n > N$.
So, it follows that $|f(y) - f(x)| = \left|f \left(x + \frac{1}{n}\right) - f(x)\right|<\epsilon$ for all $n > N$ and for all $x \in \mathbf{R}$.
As $\epsilon$ was arbitrary, we find that, for all $\epsilon > 0$, there exists an $N(\epsilon)$, such that $|f(x + 1/n) - f(x)| < \epsilon$ for all $n > N$ and for all $x \in \mathbf{R}$. Thus, $(f_n)$ converges uniformly to $f$.