Prove directly from the definition of integrability: on an interval $[a,b]$, if $f(x)$ is differentiable and $f'(x)$ is bounded, the $f(x)$ is integrable.(Do not use uniform continuity)
From the given condition, $|\lim_{x \to a}\frac{f(x)-f(a)}{x-a}|\lt B$, i.e. $|\Delta f|\lt B|\Delta x|$, for $|\Delta x|$ small.
Then how to prove it?
It suffices to show that given $\epsilon > 0$, there exists a partition $\mathcal{P}$ with $$U[f, \mathcal{P}] - L[f, \mathcal{P}] < \epsilon$$
Let $K > 0 $ be such that $|f'(x)| \leq K$ for all $x \in [a, b]$. Let $\mathcal{P}= \{x_0 = a < x_1 < \dots <x_n = b\}$ be a partition of $[a, b]$ such that $x_k - x_{k-1} < \frac{\epsilon}{K(b-a)}$ for $k = 1,\dots,n$. As each $[x_{k-1}, x_k]$ is a closed and bounded interval, $f$ attains a minimum and a maximum on this interval. Let $f$ attain its maximum at $t_k$ and minimum at $s_k$. By the mean value theorem, there exists $c$ in the interval between $t_k$ and $s_k$ with $$ f(t_k) - f(s_k) = |f'(c)(t_k - s_k)| \leq |f'(c)|(x_k - x_{k-1}) < K\frac{\epsilon}{K(b-a)} = \frac{\epsilon}{b-a}$$
$$U[f, \mathcal{P}] - L[f, \mathcal{P}] = \sum_{k=1}^n (f(t_k) - f(s_k))(x_k - x_{k-1}) < \frac{\epsilon}{b-a} \sum_{k=1}^n (x_k - x_{k-1}) = \epsilon$$