show function has a uniformly convergent subsequence

alright, so I have this question from my analysis class and I believe I have answered it correctly. I would be grateful if you could read it and give me your thoughts.

A sequence $f_n$ of real valued functions in [0,1] satisfies $f_n$$(0) = 0$ and |$f_n(x) - f_n(y)$| $<= |x-y|$. show that it has a uniformly convergent subsequence.

we were given in class the Arzela-Asocili theorem: that if a sequence of functions is equibounded, equicontinuous, and $f_n \in C(K)$ where K is a compact metric space, then there exists a uniformly convergent subsequence $f_n -> f \in C(K)$

i) We have that [0,1] is a closed, bounded subset of R, so it must be compact.

ii) to prove equiboundedness: since $f_n(0) = 0$ implies $|f_n(x)| <= |x|$ .

||$f_n$|| = sup$_x f_n(x)$ <= $sup_x|x|$ <= 1

so we have shown that each function in the sequence is bounded by one, therefore $f_n$ is equibounded.

iii) $f_n$ is equicontinuouus: since $\forall \epsilon, \exists \delta$such that $d(x,y) <= \delta$ implies $|f_n(x) - f_n(y) | <= \epsilon$. if we simply choose $\delta = \epsilon$ then since our outputs are always bounded by our inputs, the epsilon inequality will be satisfied.

therefore we have shown our 3 conditions have been met, so there must exist a convergent subsequence of $f_n$, $f$ $\in C(K)$

thats what I got. What do you guys think?