How to prove this following $\{f_n\}$ is pointwise bounded?

Question

Question. Let $\{f_n\}$ be a sequence of functions which are continuous over $[0,1]$ and continuously differentiable in $[0,1]$. Assume that $|f_n(x)|\le1$ and that $|f_n'(x)|\le 1$ for all $x \in (0,1)$ and for each positive integer $n$. Then prove that $\{f_n\}$ contains a subsequence which converges in $C[0,1]$.

My Solution: Here I try to use the following theorem from Baby Rudin:

Theorem. If $K$ is compact, if $f_n \in C(K)$ for $n = 1,2,3,\dots$ and if $\{f_n\}$ is pointwise bounded and equicontinuous o $K$, then

1. $\{f_n\}$ is uniformly bounded on $K$.

2. $\{f_n\}$ contains a uniformly convergent subsequence.

Now using MVT and the given conditions I have managed to prove that $\{f_n\}$ is equicontinuous on $[0,1]$. Now I have to prove that $\{f_n\}$ is pointwise bounded. Now $|f_n(x)|\le 1$ for all $x \in (0,1)$. So I have to prove that $\{f_n(0)\}_n$ and $\{f_n(1)\}_n$ are bounded. How can I proceed now to prove these?

Answer 1

Since $f_n$ is continuous in $[0,1]$ (and so is $|f_n|$), then $$ |f_n(0)| = \lim_{x\to 0} |f_n(x)| \leq 1, $$ and the same holds for $|f_n(1)|$.

Answer 2

This answer was posted before a correction was made by the OP. I don't know if you mis-typed the question but it is false as stated. Example, $f_n(x)=x^{n}$. However, if the hypothesis includes the condition $|f_n'(x)| \leq 1$for all $n$ and $x$ then you argument works: just use $f_n(\frac 1 2)-f_n(0)=\int_0^{1/2} f_n'(t) \, dt$ to prove that $\{f_n(0)\}$ is bounded. Similarly, for $\{f_n(1)\}$.