For $n \geq 1$, let $f_{n}(x) = x e^{-nx^2}$, $x \in \Bbb{R}$; Then the sequence $\{f_{n}\}$ is Uniformly convergent on $\Bbb{R}$?.
I did this $f_{n}(x) \rightarrow 0$ as $n \rightarrow \infty$ for each $x \in \Bbb{R}$.
I now check for $\| f_{n} - f\| = \sup_{x \in \Bbb{R}}|f_{n}(x) - f(x)| = \sup_{x \in \Bbb{R}}|f_{n}(x) - 0| = \sup_{x \in \Bbb{R}} |f_{n}(x)|$.
Maximum for $f_{n}(x)$ occurs at $x = \pm \frac{1}{\sqrt{2n}}$ whose value is $\frac{e^{-\frac{n}{2}}}{\sqrt{2n}}$.
and $\lim_{n \rightarrow \infty} \frac{e^{-\frac{n}{2}}}{\sqrt{2n}} \rightarrow 0$.
So I think $\{f_{n}\}$ is Uniformly convergent.
Now is the sequence $\{f_{n}\}$ bounded? and is it only uniformly convergent on the compact subsets of $\Bbb{R}$?
What you've proved is that, in effect, $d_\infty(f_n,f) \to 0$ in the whole real line (in particular this works for each subset of $\mathbb{R}$, compact or not). As for boundedness, since
$$ d_\infty(f_n,f_m) \leq d_\infty(f_n,f) + d_\infty(f,f_m) $$
By convergence, there exists $k \geq 1$ such that $d_\infty(f_j,f) \leq 1$ when $j \geq k$. Hence if
$$ M = max\{d_\infty(f_1,f)\dots d_\infty(f_k,f), 1\} $$
we have that $d_\infty(f_n,f_m) \leq 2M$. Note also that this same argument proves that in any metric space, convergent sequences are bounded.