Prove $p_n(x) \rightarrow \sqrt{x}$ uniformly as $n \rightarrow \infty$ 
Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_{n+1}(x) = p_n(x) + \frac{1}{2}(x - p_n(x)^2)$.  Prove that $p_n(x) \rightarrow \sqrt{x}$ uniformly for $0 \leq x \leq 1$ as $n \rightarrow \infty$.

I'm not sure where to go with this.  I can show inductively that $\forall n,\  0 \leq p_n(x) \leq 1$ for $0 \leq x \leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me.  I know that I need to show that
$\forall \epsilon>0: \exists N \in \mathbb{N}: \forall n \geq N: \forall x\in [0,1]: \big|p_n(x) - \sqrt{x}\big| < \epsilon$
Edit:  I just had a thought.  I'm given a version of Dini's theorem that reads

Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that:
(i) $f_n \rightarrow f$ pointwise on $X$, and
(ii) $f_n(x) \geq f_{n+1}(x) $ for all $x \in X, n \in \mathbb{N}$.
Then: $f_n \rightarrow f$ uniformly on $X$.

Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - \sqrt{x}$?  Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued.  The only part I would need to prove is (i), which is where I get confused.  Correct me if I'm wrong, but after that I could say that since $(-p_n)_n \rightarrow -\sqrt{x}$, then $(p_n)_n \rightarrow \sqrt{x}$.  Any advice?
 A: By the recursion formula, we have
$$p_{n + 1}(x) - \sqrt{x} = (p_n(x) - \sqrt{x})\left[1 - \frac{1}{2}(p_n(x) + \sqrt{x})\right].$$
Use this recursion $n$ times, it follows that
$$p_{n}(x) - \sqrt{x} = (p_0(x) - \sqrt{x})\prod_{k = 0}^{n - 1}\left[1 - \frac{1}{2}(p_k(x) + \sqrt{x})\right]. \tag{1}$$
We shall show that for each $k$, it holds that
$$0 \leq 1 - \frac{1}{2}(p_k(x) + \sqrt{x}) \leq 1 - \frac{1}{2}\sqrt{x}. \tag{2}$$
It is easy to see that $p_n(x) \geq 0$ for $x \in [0, 1]$ (as you stated, it in fact holds that $0 \leq p_n(x) \leq 1$, which can be proved inductively), hence the right inequality of $(2)$ holds. To show the left side inequality, notice that
\begin{align}
& 2 - p_k(x) - \sqrt{x} \\
= & 2 - p_{k - 1}(x) - \frac{1}{2}x + \frac{1}{2}p_{k - 1}^2(x) - \sqrt{x} \\ 
= & \frac{1}{2}(p_{k - 1}(x) - 1)^2 + \frac{3}{2} - \frac{1}{2}x - \sqrt{x} \\
\geq & \frac{1}{2}(p_{k - 1}(x) - 1)^2 \geq 0.
\end{align}
Therefore $(2)$ holds. Consequently, we can obtain an upper bound of the right hand side of $(1)$:
\begin{align}
& |p_n(x) - \sqrt{x}| \\
= & \sqrt{x} \prod_{k = 0}^{n - 1}\left| 1 - \frac{1}{2}(p_k(x) + \sqrt{x})\right| \\
\leq & \sqrt{x}\left(1 - \frac{1}{2}\sqrt{x}\right)^n \tag{3} 
\end{align}
Since the right side of $(3)$ converges to $0$ uniformly as $n \to \infty$, the result follows.
