$f_n(x)=n(f(x+\frac{1}{n})-f(x))$ converges uniformly to derivative 
Let $f$ be a continuous function with continuous derivative on $(a,b)\subseteq\mathbb{R}$.
Define $f_n(x)=n(f(x+\frac{1}{n})-f(x))$.
Prove that $f_n$ converges uniformly to $f'$ on any interval $[c,d]\subseteq(a,b)$.

My proof:
Let $\epsilon>0$.
Since $f'$ is continuous on $[c,d]$ it is uniformly continuous, thus there is $\delta>0$ s.t. $|f'(x)-f'(y)|<\epsilon$ whenever $|x-y|<\delta$.
By MVT,  $f_n(x)=n(f(x+\frac{1}{n})-f(x))=f'(t)$ for some $t\in(x,x+\frac{1}{n})$.
Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\delta$. Then $\forall n\ge N$, we have:
$\sup\limits_{x\in[c,d]}|f_n(x)-f'(x)|=\sup\limits_{x\in[c,d]}|f'(t)-f'(x)|<\epsilon$, by uniform continuity, so done.

Can anyone please check if my proof above is correct? Thank you.
 A: I think there is one small problem in your proof. We can't be sure that the point $t$ is still in $[c,d]$. For example in you take $x=d$ then $x+\frac{1}{n}$ will always be outside of the interval $[c,d]$, hence $t$ will be outside of the interval as well. And outside the interval you can't be sure the same $\delta$ works. And even if we forget about the point $x=d$ itself, other points still cause problems: if you take $x$ which is just a bit smaller than $d$ then you might need a very large $n$ to make sure that $x+\frac{1}{n}\in [c,d]$. And when points depend on $n$ it is a problem, because you need uniform convergence here. 
Anyway, there is a simple way to fix the problem: find a big enough $M$ such that $d+\frac{1}{M}<b$, and then choose $\delta$ by uniform continuity in the interval $[c,d+\frac{1}{M}]$. From this moment it will be fine, because for each $n>M$ and for each $x\in [c,d]$ we indeed have $x+\frac{1}{n}\in [c,d+\frac{1}{M}]$. In this case all the points $t$ will indeed satisfy $|f'(t)-f'(x)|<\epsilon$ because $\delta$ works everywhere in $[c,d+\frac{1}{M}]$ while you take values of $x$ only from $[c,d]$. 
