Prove $p_n(x) \rightarrow \sqrt{x}$ uniformly as $n \rightarrow \infty$

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.