Stuck on proof of step 4 on Thm 7.32 on baby Rudin. According to baby Rudin's thm 7.32
Then the uniform closure $B$ of $A$ consists of 'all' real continuous functions of $K$.
and Step 4 in proof:
Given a real function $f$, continuous on $K$, and $\epsilon>0$ there exists a function $h\in B$ such that
$$|h(x)-f(x)|<\epsilon\ \ \ (x\in K)$$
"Since $B$ is uniformly closed, this statement is equivalent to the conclusion of the theorem."
I think we can get continuity if above is satisfied, because of following
$$|h(x)-h(t)|\leq |h(x)-f(x)|+|f(x)-f(t)|+|f(t)-h(t)|<3\epsilon,\ \ \ (x,t\in K)$$
$$\because f \ is \ continuous \ on\ K $$
Then why the uniformly closed is relevant with "all"?
Please help me!!!

The statement of part 4 can be through of like this. Given any real continuous function $f$ on $K$, we can find a sequence of $h_{n}\in B$ such that $h_{n}\to f$ uniformly. 
Uniformly closed means that for any uniformly convergent sequence $h_{n}\in B$, $\lim_{n\to\infty} h_{h}\in B$. Thus, we conclude via part 4 that for any real contionus function $f$ on $K$, $f\in B$, which is the desired result.
