We have a sequence $(f_n)$ on $[0,\infty)$, defined by $f_n(x)=\frac{\cos(nx)}{e^{(nx)}}$. The limit function $(f_n)$ of this sequence is $0$ for $x>0$ and $1$ for $x=0$.
First part of the question ask if this sequence converges uniformly on $x\in [0,\infty)$.
As it is not continuous, and it has a "jump" at $x=0$, I would say it is not, as all uniformly convergent sequences have a continuous limit function. Is this correct?
For the second part, we have to prove that it converges uniformly on $\forall$ sets $A$, where $\inf A>0$ and $x\in A$.
How do I do that? I know that $|f_n(x)-f_n| \leq \frac{1}{e^{nx}}$. But now I do not know, how to continue.
For the first part, yes, this is basically* correct (*you should add one phrase: "all uniformly convergent sequences of continuous functions have a continuous limit function").
For the second part, let $a = \inf A > 0$. We show that $f_n \to \mathbb{1}_{\{0\}}$ uniformly on $[a, \infty)$, where the limit function is the indicator function $\mathbb{1}_{\{0\}}(x)$ which is equal to $1$ if $x \in \{0\}$ and $0$ if $x \notin \{0\}$. Hence $$ \sup_{x \in [a, \infty)}\left| f_n(x) - \mathbb{1}_{\{0\}}(x) \right| = \sup_{x \in [a, \infty)} \left| f_n(x) \right| \leq \sup_{x \in [a, \infty)} e^{-nx}. $$ Now the point is that $e^{-nx}$ is monotonically decreasing, so its largest value occurs at the smallest value of $x$, which is $a$; hence $$ \sup_{x \in [a, \infty)}\left| f_n(x) - \mathbb{1}_{\{0\}}(x) \right| \leq \sup_{x \in [a, \infty)} e^{-nx} = e^{-n a}. $$ If we pick $N$ so large that $e^{-Na} < \varepsilon$, then we have $n \geq N \implies \sup_{x \in [a, \infty)}\left| f_n(x) - \mathbb{1}_{\{0\}}(x) \right| < \varepsilon$, as desired, so that $f_n \to \mathbb{1}_{\{0\}}$ uniformly on $[a, \infty)$ for any $a > 0$. Now any $A \subset [0, \infty)$ with $\inf A = a$ is a subset of $[a, \infty)$, and hence $f_n \to \mathbb{1}_{\{0\}}$ uniformly on $A$ also.
(This is exactly what @G.Sassatelli already said, but with more words, which help to shed light on what's going on.)