Question:
$\{f_n(x)\}$ is continuous on $[a,b]$ and has pointwise convergence on $[a,b]$ (to $f(x)$) .
$\forall \epsilon>0,\exists\delta>0,\forall x,y\in[a,b](|x-y|<\delta),\forall n\geqslant 1,|f_n(x)-f_n(y)|<\epsilon$ .
Prove : $\{f_n(x)\}$ uniformly converges on $[a,b]$ .
Standard Proof:
1.Prove $f(x)$ is uniformly continuous.
2.$|f_n(x)-f(x)|\leqslant |f_n(x)-f_n(y)|+|f_n(y)-f(y)|+|f(y)-f(x)|$ .
3.Use Heine-Borel theorem to deal with $|f_n(y)-f(y)|$ ( $\cup_{y\in[a,b]}(y-\delta_y,y+\delta_y)$ overlaps $[a,b]$ ).
My 'Proof':
$\forall \epsilon>0,\exists\delta>0,\forall x,y\in[a,b](|x-y|<\delta),\forall n\geqslant 1,|f_n(x)-f_n(y)|<\epsilon$ .
1.Use $\delta$ to divide $[a,b]$ equally,so that $\forall x\in[a,b]$ , $x$ must be in $[y_k,y_{k+1}]$ and thus $|x-y_k|<\delta$:
$M=[\frac{1}{\delta}]+1,y_k=a+\frac{k}{M}(b-a),k=0,1,...,M$ .
$\forall y_k,\exists N_k>0,\forall n>N_k,|f_n(y_k)-f(y_k)|<\epsilon$ .
$N=\max\limits_{0\leqslant k\leqslant M} N_k,\forall n,m>N,\forall x\in[a,b],\exists y_k(|x-y_k|<\delta)$ ,
2.Estimate $|f_n(x)-f_m(x)|$ :
$|f_n(x)-f_m(x)|\leqslant |f_n(x)-f_n(y_k)|+|f_n(y_k)-f(y_k)|+|f(y_k)-f_m(y_k)|+|f_m(y_k)-f_m(x)|<4\epsilon,$
the first and fourth part is less than $\epsilon$ due to the equicontinuity,the second and third part is less than $\epsilon$ because we have picked $N$(which does not rely on $x$).
3.Take the sup(n,m is defined before $x$):
$\sup\limits_{x\in[a,b]}|f_n(x)-f_m(x)|<4\epsilon$ .
Am I correct?Thanks for your checking!
Addition:
Is taking $N=\max\limits_{0\leqslant k\leqslant M} N_k$ legal?(for the maximum is well-defined only when $M$ is finite.)
In other words,does the estimation 'always' hold?
Yes, it's correct. Actually your answer is quite same with the standard proof. The key ideas are both that the covergence on any point $x$ can be controlled by the convergence on a nearby point $y_k.$ So the convergence on $x\in [a,b]$ can be concluded to the convergences on $\{y_k\}_{k=1}^M,$ while the latter one is easy to control.
The only difference is that you use Cauchy sequence to prove the convergence, while the standard proof proves directly.