Theorem. If $f:[a,b]\to\mathbb{R}$ be a Riemann integrable function on $[a,b]$ then show that for all $\varepsilon>0$ there exists $\delta(\varepsilon)>0$ such that for all partition $P_\varepsilon$ with $\lVert P_\varepsilon\rVert<\delta(\varepsilon)$ we will have $U(f;P_\varepsilon)-L(f;P_\varepsilon)<\varepsilon$.
Proof. The proof of the part that for all $\varepsilon>0$ there exists partition $P_\varepsilon$ such that $U(f;P_\varepsilon)-L(f;P_\varepsilon)<\varepsilon$ is easy and will not be discussed here. I focus mainly on the existence of $\delta(\varepsilon)$.
Fix $\varepsilon>0$ and observe that for all $\varepsilon>0$ there exists $P_\varepsilon$ and $Q_\varepsilon$ such that, $$U(f)\le U(f;P_\varepsilon)<U(f)+\varepsilon$$$$L(f)\ge L(f;Q_\varepsilon)>L(f)-\varepsilon$$where $U(f)$ and $L(f)$ are respectively upper and lower Darboux intergrals of $f$ on $[a,b]$.
Now consider the sets, $$\mathcal{U}\langle\varepsilon\rangle:=\{\lVert P_\varepsilon\rVert:U(f)\le U(f;P_\varepsilon)<U(f)+\varepsilon\}$$ $$\mathcal{L}\langle\varepsilon\rangle:=\{\lVert Q_\varepsilon\rVert:L(f)\ge L(f;Q_\varepsilon)>L(f)-\varepsilon\}$$Then observe that both $\sup\mathcal{U}\langle\varepsilon\rangle$ and $\sup\mathcal{L}\langle\varepsilon\rangle$ exists (because both sets are bounded above by $(b-a)$) and they are positive since $P_\varepsilon, Q_\varepsilon$'s are partitions. We choose, $$0<\delta(\varepsilon)<\min\left(\sup\mathcal{U}\langle\varepsilon\rangle,\sup\mathcal{L}\langle\varepsilon\rangle\right)$$Now observe that for all $P_\varepsilon$ such that $\lVert P_\varepsilon\rVert<\delta(\varepsilon)$, $\lVert P_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$ (such $P_\varepsilon$'s exist because if $\lVert P_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle$ and $\lVert Q_\varepsilon\rVert\in \mathcal{L}\langle\varepsilon\rangle$ then $\lVert P_\varepsilon\cup Q_\varepsilon\rVert\in \mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$).
This completes the proof.
Questions
Is my proof of the theorem correct?
Is there any part of the written proof that isn't explained clearly?
The proof is incorrect. There is circular reasoning to the extent that partitions in the class $\mathcal{U}\langle\varepsilon\rangle\cap\mathcal{L}\langle\varepsilon\rangle$ already satisfy the desired inequality. You need to show that you can produce $\delta(\epsilon)$ such that the inequality is satisfied for any partition with $\lVert P \rVert < \delta(\epsilon).$
Your argument implicitly uses the assumption that, for any partition $P,$ the condition $\lVert P \rVert < \lVert P_\epsilon \rVert$ implies that $U(P,f) - L(P,f) < U(P_\epsilon,f) - L(P_\epsilon,f).$ This would be true if $P$ were a refinement of $P_\epsilon$, but not simply because the norm is smaller. Recall that the norm of a partition is the length of the longest sub-interval.
For a counterexample to the assumption, consider the Riemann integrable function $f:[0,1] \to \mathbb{R}$ where
$$f(x) = \begin{cases} 0, \,\,\, 0 \leqslant x < 1/2 \\1 , \,\,\, 1/2 \leqslant x \leqslant 1 \end{cases}$$
Consider the partitions, $P_\epsilon = (0,1/2- \delta, 1/2+\delta,1)$ and $P = (0,1/2- \delta', 1/2+\delta',1).$
Choose $\delta < 1/6$ and $\delta' = 1/6.$
Then
$$\lVert P_\epsilon \rVert = \max(2\delta, 1/2 - \delta) > 1/3, \\ \lVert P \rVert = \max(2\delta', 1/2 - \delta') = 1/3$$
and
$\lVert P \rVert < \lVert P_\epsilon \rVert ,$ but
$$U(P_\epsilon,f) - L(P_\epsilon,f) = 2\delta < 1/3, \\ U(P,f) - L(P,f) = 2\delta' = 1/3,$$
and $U(P,f) - L(P,f) > U(P_\epsilon,f) - L(P_\epsilon,f).$
Valid Proof
Suppose $f$ is Riemann integrable with integral equal to $I$. Given any $\epsilon >0$ there exists a partition $P_\epsilon$ with $U(P_\epsilon,f) < I + \epsilon/4$ and $L(P_\epsilon,f) > I - \epsilon/4.$
Let $D = \sup\{|f(x)-f(y)|:x,y \in [a,b] \}$ denote the maximum oscillation of $f$ and let $\delta = \epsilon/(4mD)$ where $m$ is the number of points in the partition $P_\epsilon$.
Now let $P$ be any partition with $\lVert P \rVert<\delta$. Form the common refinement $Q = P \cup P_\epsilon$.
Since $Q$ is a refinement of $P_\epsilon$ we have
$$L(P_\epsilon,f) < L(Q,f) \leqslant U(Q,f) < U(P_\epsilon,f).$$
We can see that terms of the upper sums $U(P,f)$ and $U(Q,f)$ differ in at most $m$ sub-intervals where the deviation is bounded by $\delta D$ and
$$U(P,f) < U(Q,f) + m D \delta = U(Q,f) + \epsilon/4.$$
It follows that
$$U(P,f) < U(Q,f) + \epsilon/4 < U(P_\epsilon,f) + \epsilon/4 < I + \epsilon/2.$$
By a similar argument, we can show $L(P,f)> I - \epsilon/2$.
Hence, $U(P,f)-L(P,f) < \epsilon$.