By the definition of $W^{1,p}$, there exist $v_n \in C_{c}^{1}(\mathbb{R})$ and $w \in L^p(\mathbb{R})$ such that $v_n \to u$ in $L^p(\mathbb{R})$ and $v_n' \to w$ in $L^p(\mathbb{R})$. In this case we write $u'=w$. Note that $v_n'$ is a classical derivative, so $$D_h v_n(x)= I_h(v_n')(x) \,, \tag{1} $$ where for $f\in L^p(\mathbb{R})$, we write $$I_h(f)(x):=\int_0^h \frac{f(x+t)}{h} \,dx \,.$$ By Jensen's inequality [1], for all $n \ge 1 $ and $x \in \mathbb{R}$, we have $$|I_h(v_n')(x)-I_h(u')(x)|^p \le \frac{1}{h} \int_0^h |v_n'(x+t)-u'(x+t)|^p \,dt \,. $$ Integrating both sides $\,dx$ and using Fubini on the right-hand side, we obtain $$\|I_h(v_n') -I_h(u') \|^p \le \frac{1}{h}\int_0^h \|v_n'(\cdot+t)-u'(\cdot+t)\|_p^p \,dt= \|v_n' -u' \|_p^p \,.\tag{2} $$

Given $\epsilon>0$, find $k$ such that $$ \|v_k'-u'\|_p<\epsilon \,. \tag{3} $$ Let $M$ denote the Lebesgue measure of the support of $v_k$. Since $v_k'$ is uniformly continuous, there exists $h_0\in(0,1)$ such that $$\forall t\in [0, h_0], \quad \sup_{x \in \mathbb{R}} |v_k'(x+t)-v_k'(x)|<\epsilon/(M+1) \,,$$ so for $h\in [0, h_0]$ and all $x$, we have $|I_h (v_k')(x)-v_k'(x)|<\epsilon/(M+1)$, whence $$\|I_h (v_k') -v_k'\|_p^p \le (M+h) (\epsilon/(M+1))^p <\epsilon^p \,.$$ In conjunction with $(2)$ and $(3)$, this gives $$\|I_h(u')-u' \|_p \le \|I_h(u')-I_h(v_k') \|_p + \|I_h(v_k')-v_k' \|_p + \| v_k' -u'\|_p <3\epsilon \,. \tag{4}$$

next, fix $h\in [0, h_0]$, and choose $m=m(h,\epsilon)$ such that $$\|u-v_m \|_p<\epsilon h \quad \text{and} \quad \|u'-v_m'\|_p<\epsilon \,. $$ The first inequality implies that $\|D_h(u) -D_h(v_m) \|_p<2\epsilon$. Therefore, by $(1),\, (2)$ and $(4)$, $$ \|D_h(u)-u'\|_p \le \|D_h(u) -D_h(v_m) \|_p+\|I_h(v_m')-I_h(u')\|_p+\|I_h(u')-u' \|_p < 2\epsilon+\epsilon+3\epsilon=6\epsilon \,. $$ This completes the proof.

[1] https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form

The proof is not short, because it is done from first principles, without using any theorems about Sobolev space except its definition.

By the definition of $W^{1,p}$, there exist $v_n \in C_{c}^{1}(\mathbb{R})$ and $w \in L^p(\mathbb{R})$ such that $v_n \to u$ in $L^p(\mathbb{R})$ and $v_n' \to w$ in $L^p(\mathbb{R})$. In this case we write $u'=w$. Note that $v_n'$ is a classical derivative, so $$D_h v_n(x)=\frac{v_n(x+h)-v_n}{h}= I_h(v_n')(x) \,, \tag{1} $$ where for $f\in L^p(\mathbb{R})$, we write $$I_h(f)(x):=\int_0^h \frac{f(x+t)}{h} \,dx \,.$$ By Jensen's inequality [1], for all $n \ge 1 $ and $x \in \mathbb{R}$, we have $$|I_h(v_n')(x)-I_h(u')(x)|^p \le \frac{1}{h} \int_0^h |v_n'(x+t)-u'(x+t)|^p \,dt \,. $$ Integrating both sides $\,dx$ and using Fubini on the right-hand side, we obtain $$\|I_h(v_n') -I_h(u') \|^p \le \frac{1}{h}\int_0^h \|v_n'(\cdot+t)-u'(\cdot+t)\|_p^p \,dt= \|v_n' -u' \|_p^p \,.\tag{2} $$

Given $\epsilon>0$, find $k$ such that $$ \|v_k'-u'\|_p<\epsilon \,. \tag{3} $$ Let $M$ denote the Lebesgue measure of the support of $v_k$. Since $v_k'$ is uniformly continuous, there exists $h_0\in(0,1)$ such that $$\forall t\in [0, h_0], \quad \sup_{x \in \mathbb{R}} |v_k'(x+t)-v_k'(x)|<\epsilon/(M+1) \,,$$ so for $h\in [0, h_0]$ and all $x$, we have $|I_h (v_k')(x)-v_k'(x)|<\epsilon/(M+1)$, whence $$\|I_h (v_k') -v_k'\|_p^p \le (M+h) (\epsilon/(M+1))^p <\epsilon^p \,.$$ In conjunction with $(2)$ and $(3)$, this gives $$\|I_h(u')-u' \|_p \le \|I_h(u')-I_h(v_k') \|_p + \|I_h(v_k')-v_k' \|_p + \| v_k' -u'\|_p <3\epsilon \,. \tag{4}$$

Next, fix $h\in [0, h_0]$, and choose $m=m(h,\epsilon)$ such that $$\|u-v_m \|_p<\epsilon h \quad \text{and} \quad \|u'-v_m'\|_p<\epsilon \,. $$ The first inequality implies that $\|D_h(u) -D_h(v_m) \|_p<2\epsilon$. Therefore, by $(1),\, (2)$ and $(4)$, \begin{eqnarray} \|D_h(u)-u'\|_p &\le& \|D_h(u) -D_h(v_m) \|_p+\|I_h(v_m')-I_h(u')\|_p+\|I_h(u')-u' \|_p \\ &<& 2\epsilon+\epsilon+3\epsilon=6\epsilon \,. \end{eqnarray} This completes the proof.

[1] https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form