What is the main difference between pointwise and uniform convergence as defined here? I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.
Let $E$ be a non-empty subset of $\Bbb{R}$. A sequence of functions $\{f_n\}_{n\in \Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
On the other hand $\{f_n\}_{n\in \Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
QUESTION:
Why is $f_n(x)\to f(x),\;\forall\,x\in E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?

Sorry, but yes, you probably are missing something important, because the second statement in your post

On the other hand $\{f_n\}_{n\in\mathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
  $$f_n(x)\to f(x),\;\forall\,x\in E.$$

is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.
Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $\varepsilon/\delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.
Your best bet is to check the original source to find out what exactly it says there.
