I know that if function is uniformly convergent ($ |f_n(x)-f(x)|<\epsilon. \forall n > N(\epsilon)$), it is Cauchy convergent ($ |f_n(x)-f_m(x)|<\epsilon. \forall n,m > N(\epsilon)$)
So my question is: if sequence is Cauchy convergent does this imply uniform convergence? I think the answer is no, but can't figure out example.
Yes it does; suppose $\{f_n\}$ is a series of functions on a subset $E$ of $\mathbb{R}$ such that for all $\epsilon > 0$ there exists a $N(\epsilon)$ such that $m, n > N(\epsilon)$ implies that $|f_m(x) - f_n(x)| < \epsilon$ for all $x \in E$.
Since $f_n(x)$ is a Cauchy sequence for each particular $x \in E$, we know that our sequence $f_n(x)$ has a pointwise limit; we call it $f(x)$. We must now show that the convergence to $f(x)$ is uniform.
So let $\epsilon > 0$. By our assumption on $\{ f_n \}$ there exists an $N(\epsilon)$ such that $m, n > N(\epsilon)$ implies that $$|f_m(x) - f_n(x)| < \epsilon$$ for all $x \in E$. Now comes the interesting step: fix $n$, and take the limit $m \to \infty$ in the above expression. The result is that $$|f(x) - f_n(x)| \leq \epsilon$$ for all $x \in E$ and $n > N(\epsilon)$. This proves the uniform convergence.
(This last step has to be justified; it relies on the fact that the function $\phi(x) = |x - c|$ is continuous.)