I have trouble in proving the following exercise:
Let $\omega(f,x)<\epsilon$ for every $x\in[a,b].$ Show there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta.$ (Hint: Use the fact that $[a,b]$ is compact.) We assume that $f:[a,b]\to\mathbb{R}$ is bounded on $[a,b].$ Here $\omega(f,x)$ is the oscillation of $f$ at point $x.$ That is, \begin{gather*} \omega(f,x)=\lim_{\delta\to 0^+}\Big(\sup\{f(t)\mid t\in (x-\delta,x+\delta)\cap[a,b]\}- \inf \{f(t)\mid t\in (x-\delta,x+\delta)\cap[a,b]\}\Big). \end{gather*}
This exercise comes from Exercise 1.2.4 of Rana's book An Introduction to Measure and Integration, 2e.
I have tried the following: Let $\mathbb{B}(x_0,\eta)$ denote the neighborhood of $x_0$ with radian $\eta.$ That is, $\mathbb{B}(x_0,\eta):=(x_0-\eta,x_0+\eta).$ Because $0<\epsilon/2<\epsilon,$ and, by hypothesis, $$\omega(f,x):=\lim\limits_{\delta\to0^+}\omega(f,\mathbb{B}(x,\delta)\cap [a,b])<\epsilon $$ for every $x\in[a,b],$ by order preserving of limit, for each $x\in[a,b],$ there exists a $\delta(x)>0$ such that $\omega(f,\mathbb{B}(x,\delta(x))\cap [a,b])<\epsilon/2,$ which implies that for every $y\in\mathbb{B}(x,\delta(x))\cap [a,b],$ then $|f(x)-f(y)|<\epsilon.$ Because $[a,b]\subset \cup_{x\in[a,b]} \mathbb{B}(x,\delta(x)),$ and $[a,b]$ is compact, there exist $x_1,\dots, x_n\in[a,b],$ such that $\cup_{j=1}^n \mathbb{B}(x_j,\delta(x_j))\supset [a,b].$ Let $\delta=\min\{\delta(x_j)\mid j=1,\dots,n\}.$
Since $[a,b]\subset\cup_{x\in[a,b]} \mathbb{B}(x,\delta),$ there exist $y_1,\dots, y_k\in [a,b]$ such that $[a,b]\subset \cup_{j=1}^k\mathbb{B}(y_j,\delta).$
But I do not know how to go on? Can anyone help me?
PS: After some consideration, I found the Lebesgue number is important here. So I rewrite my previous proof as follows:
Note that $\omega(f,x)=\lim_{r\to 0^+}\omega(f,\mathbb{B}(x,r)\cap [a,b]).$ Put $I:=[a,b].$ Because $\omega(f,x)<\frac{\omega(f,x)+\epsilon}{2}<\epsilon,$ by definition of $\omega(f,x)$ and local order-preserving of limit, for every $x\in I,$ there exists $\eta(x)>0$ such that for all $0<\xi<\eta(x),$ $\omega(f,\mathbb{B}(x,\eta)\cap I)<\frac{\omega(f,x)+\epsilon}{2}.$ Particularly, set $\xi(x)=\eta(x)/2,$ and so $\omega(f,\mathbb{B}(x,\xi(x))\cap I)<\frac{\omega(f,x)+\epsilon}{2}.$ Clearly $I\subset \cup_{x\in I} \mathbb{B}(x,\xi(x)).$ As $I$ is compact, we can extract a finite number of points, for example, $x_1,\dots, x_n,$ such that $I\subset \cup_{j=1}^n \mathbb{B}(x_j,\xi(x_j)).$ Put $\zeta_j=\xi(x_j)=\eta(x_j)/2$ for every $j=1,\dots,n.$ Let $l$ be the Lebesgue number for the covering $\{B(x_j,\zeta_j)\mid j=1,\dots, n\}$ of $I.$ Now take $$\delta=\min\{l/2, \zeta_1/2,\dots, \zeta_n/2\}.$$ For every $x, y\in I,$ if $|x-y|<\delta,$ then, $|x-y|<l,$ and so, by definition of the Lebesgue number, there exists some neighborhood $\mathbb{B}(x_k,\zeta_k)$ in the foregoing covering $\{B(x_j,\zeta_j)\mid j=1,\dots, n\},$ such that $\{x,y\}\subset \mathbb{B}(x_k,\zeta_k).$ Hence, \begin{align*} |f(x)-f(y)|\leq \omega(f,\mathbb{B}(x_k,\zeta_k)\cap I)<\frac{\omega(f,x)+\epsilon}{2}<\epsilon. \end{align*}
For each $x\in[a, b] $ let $\epsilon_x=(\omega(f, x) +\epsilon) /2$ and then it is obvious that we have a neighborhood $B(x, r) $ such that $|f(u) -f(v) |<\epsilon_x$ whenever $u, v\in B(x, r) $. By compactness of $[a, b] $ a finite number of neighborhoods of type $B(x, r) $, say $B(x_i, r_i) $ for $i=1,2,\dots, n$, cover the interval $[a, b] $. The end points of these neighborhoods which lie in interval $[a, b] $ form a partition and if $\delta$ is the norm of this partition then our $\delta$ works as asked in question. If we choose any two points $x, y$ such that $|x-y|<\delta$ then both these points lie in same neighborhood $B(x_k, r_k) $ and hence $|f(x) - f(y) |<\epsilon_{x_k} <\epsilon $.
Your usage of $\epsilon/2$ is wrong. If something has a limit less than $\epsilon$ then you can not say that its values are less than $\epsilon/2$. Rather you can say that the values are closer to the limit than to $\epsilon$ and thus if the limit is $L<\epsilon$ then you can ensure the values to be less than $L'$ where $L'$ is any chosen number such that $L<L'<\epsilon$. In my answer that role of $L'$ is played by $\epsilon_x$.