Prove that $\sum_{n=1}^\infty f_n$ converges uniformly on $\mathbb{R}$ where $f_n$ is defined piecewise. 
Prove that the series of functions $\sum_{n=1}^\infty f_n$ converges
  uniformly on $\mathbb{R}$, where $$f_n: \mathbb{R} \to \mathbb{R}: x
 \mapsto \begin{cases}0 \quad x \neq n \\\frac{1}{x} \quad x
 =n\end{cases}$$

My attempt:
Let $x \in \mathbb{R}$ be fixed. Let $n \in \mathbb{N}$ with $n > x$. Then, it is clear that:
$$\sum_{k=1}^n f_k(x) = \begin{cases}0 \quad x \notin \mathbb{N_0}\\\frac{1}{x} \quad x \in \mathbb{N_0}\end{cases}$$
Hence, letting $n \to \infty$, we have that the given series converges pointwise to the function $$f: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases}0 \quad x \notin \mathbb{N_0}\\\frac{1}{x} \quad x \in \mathbb{N_0}\end{cases}$$
We now show uniform convergence. Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$.
If $x \in \mathbb{N_0}$, we consider two cases:
(i) $n\geq x$: then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = 0$
(ii) $n < x$: then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = |f(x)| = \frac{1}{x} < \frac{1}{n}$
If $x \notin \mathbb{N_0}$, then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = 0$
So, we have proven that $\forall n \in \mathbb{N}, \forall x \in \mathbb{R}: \left|\sum_{k=1}^nf_k(x) - f(x)\right| < \frac{1}{n} \to 0$
Let then $\epsilon > 0$. Choose $n_0: \forall n \geq n_0: \frac{1}{n} < \epsilon$. Then, for $n \geq n_0: \left|\sum_{k=1}^nf_k(x) - f(x)\right| < \frac{1}{n} < \epsilon$
So, the convergence is uniform. Is this correct?

You are mostly there. With the $f$ you have found as the limiting function, for any $x\in \mathbb{R}$,
$$
\left|\sum_{k=1}^nf_k(x)-f(x)\right|\leq \frac{1}{n+1}
$$
since if $x\not\in \mathbb{N}$ or if $x\in \{ 1,\dots,n\}$ the difference is zero. Otherwise, the worst this difference could be is the difference at $x=n+1$. This gives you
$$
\sup_{x\in \mathbb{R}}\left|\sum_{k=1}^nf_k(x)-f(x)\right|\leq \frac{1}{n+1}\to 0
$$
as required.
