The definition I have been given on my course for uniform convergence is that
A sequence of functions $f_n:D\to\mathbb{R}$ converges uniformly to a function $f:D\to\mathbb{R}$ if the real sequence $$||f_n - f||=\sup(|f_n-f|:x\in D)\to 0$$
I have been asked to find a sequence of discontinuous functions $f_n : [0,1]\to\mathbb{R}$ that uniformly converges to a continuous function. I chose $$f_{n}(x)=\begin{cases}
\frac{1}{n} & x=0\\
0 & x\neq0
\end{cases}$$
as my sequence and claim that it converges uniformly to the continuous (constant) function $f(x)=0$.
Proof
Let $f_n$ and $f$ be defined as above, then $$\left\Vert f_{n}-f\right\Vert =\left\Vert f_{n}-0\right\Vert
=\sup\{|f_{n}|:x\in[0,1]\}
=|\frac{1}{n}|$$
Now let $\epsilon >0$ be given, by the Archimedean property we can find $N\in\mathbb{Z^+}$ such that $N>\frac{1}{\epsilon}\implies\frac{1}{N}<\epsilon$. So choose such $N$, then for all $n\geq N$ we have $$\left\Vert f_{n}-f\right\Vert =|\frac{1}{n}|<\epsilon\implies\left\Vert f_{n}-f\right\Vert\to0$$ and thus, $f_n$ converges uniformly to $f$.
Is this a valid proof? My only concern is having chosen a constant function to converge to.