Using proof by contradiction to prove Dini's theorem

Question

Does this proof sound right? Thanks

Dini's Theorem: If $f$ and $f_n$ are continuous functions on $[a,b]$ such that $f_n \leq f_{n+1} \forall n \geq 1$ and $(f_n)$ converges to $f$ pointwise, then $(f_n)$ converges to $f$ uniformly.

My proof:

Let $g_n = f - f_n$. Fix $x_0 \in [a,b]$. Fix $\epsilon >0$ Use the pointwise convergence of $f_n$ to find an $N(x_0, \epsilon) \in \mathbb{N}$ such that $||f_k(x_0)-f(x_0)|| < \frac{\epsilon}{3} \forall k \geq N$. Since $f$ and $f_n$ are continuous, $\forall \epsilon > 0 \exists \delta(\epsilon, x_0) > 0$ such that $||f_n(x) - f_n(x_0)|| < \frac{\epsilon}{3}$ $\forall x \in [a,b]$ with $|x-x_0| < \delta$ and similarly for $f$. Then \begin{align} g_N(x) &= f(x) - f_N(x)\\ & \leq ||f(x) - f(x_0)|| + ||f(x_0)-f_N(x_0)|| + ||f_N(x_0) - f_N(x)|| \\ &\leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align}

Since $x_0$ was an arbitrary point, the function $g_n$ is uniformly convergent in every open interval around an arbitrary point $x_0 \in [a,b]$

Now suppose for contradiction that $f_n$ does not converge uniformly to $f$. $\lim_{n\to\infty}||f-f_n||_\infty = d > 0$ $\implies$ $\exists x_1 \in [a,b]$ such that $\lim_{n\to \infty}||f(x_1) - f_n(x_1)|| = d > 0$

Choose $\epsilon = \frac{d}{2}$ and choose a sequence $x_n \in [a,b]$ that converges to $x_1$.

$\lim_{n\to\infty}g_N(x_1)= f(x_1) - f_N(x_1) = d$ which is not less than epsilon and we have arrived at a contradiction.

This implies that $f_n \to f$ uniformly.

Answer 1

Couple things:

1. What's $\delta$? Please provide the exact meaning of symbols, otherwise it is confusing.
2. You know that for each $x_0$ there exists $N(x_0, \varepsilon) \in \mathbb N$, then how could you deduce that $g_n \rightrightarrows g$ nearby each $x_*$?
3. Where does the assumption $f_n \leqslant f_{n+1}$ applied?

Additional problem

4. Is it necessary that such $x_1$ exists when $\lim \Vert f- f_n \Vert_\infty = d > 0 $?