Show uniform convergence implies equicontinuity and uniform boundedness

Hi guys! I was wondering if you could look at my proof and tell me if you think its correct/rigorous enough. I'm having a little difficulty with this.

The question is: Let (f_n) be a sequence of functions in C([0,1]), with f_n uniformly converging to f in [0,1]. Show without using the Heine-Borel theorem that the sequence is also uniformly bounded and equicontinuous in [0,1].

**Equicontinuity:**

By uniform convergence, the sequence of functions meets the Caunchy criterion, so there exists a natural number N such that when n,m>=N:

|f_n(x)-f_m(x)|<E/3 (for every x in [0,1])

By continuity of each f_k, there exists a d such that:

|x-y|<d implies |f_k(x)-f_k(y)|<e/3 (for every x,y in [0,1])

Combining these two facts, we have that for any e>0, when n,k>=N and |x-y|<d:

|f_n(x)-f_n(y)| =< |f_n(x)-f_k(x)| + |f_k(x)-f_k(y)| + |f_k(y)-f_n(y)|

|f_n(x)-f_n(y)| =< e/3 + e/3 + e/3

|f_n(x)-f_n(y)| =< e (for every x,y in [0,1])

Which is exactly the definition of equicontinuity. So, f_n is equicontinuous on [0,1].

**Uniform Boundedness:**

Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

I don't think this last part works, but I'm not sure what else to do. Maybe something like:

If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

Any comments/helpful tips are much appreciated. =)