I'm afraid I don't quite understand one of the steps in the following proof, I will highlight the offending step in red. The result is Theorem 9.12 (Morrey) in Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations. I will discuss this in the one dimensional case to keep the notation clean.
Theorem: For all $u \in W^{1,2}(\mathbb{R})$, we have
$$ |u(x)-u(y)| \leq C|x-y|^\alpha \|u'\|_{L^2} \quad a.e\quad x,y \in \mathbb{R}, $$ where $\alpha=1/2$.
Proof,
We first prove this result for $u \in C_c^1(\mathbb{R})$. Let $x \in (-\frac{r}{2},\frac{r}{2})$ for some $r >0$. We have,
$$ u(x)-u(0)=\int_0^t \frac{d}{dt}u(tx)dt, $$ thus, $$ |u(x)-u(0)|\leq \int_0^t|x|\cdot|u'(tx)|dt \leq \frac{r}{2}\int_0^t|u'(tx)|dt. $$ Set, $$ \bar{u}=\frac{1}{r}\int_{-r/2}^{r/2}u(s)ds. $$ Integrating our above inequality over the interval $(-\frac{r}{2},\frac{r}{2})$, we see, \begin{align} |\bar{u}-u(0)| & \leq \frac{r}{2r}\int_{-r/2}^{r/2}\int_0^t|u'(tx)|dtdx\\ & = \frac{1}{2}\int_0^t\int_{-r/2}^{r/2}|u'(tx)|dxdt\\ & = \frac{1}{2}\int_0^t\int_{-tr/2}^{tr/2}|u'(v)|t^{-1}dvdt, \end{align} Then applying Holder's inequality to the inner integral we get, \begin{align} \int_{-tr/2}^{tr/2}|u'(v)|t^{-1}dv & \leq \left( \int_{-tr/2}^{tr/2}|u'(v)|^2dv\right)^{1/2}\left(\int_{-tr/2}^{tr/2}t^{-2}dv\right)^{1/2}\\ & \leq \sqrt{\frac{r}{t}} \cdot\|u'\|_{L_2}. \end{align} Thus we have, \begin{align} |\bar{u}-u(0)| & \leq \frac{\|u'\|_{L_2}}{2}\int_0^t\sqrt{\frac{r}{t}}dt\\ & = \sqrt{r} \cdot\|u'\|_{L_2}. \end{align}
$\color{red}{\text{By translation, this inequality remains true for all intervals $I$ of length $r$, thus we have,}}$ $$ |\bar{u}-u(x)|\leq \sqrt{r} \cdot\|u'\|_{L_2} \quad \forall x\in I. $$ $\color{red}{\text{By adding these (and using the triangle inequality) we obtain}},$ $$ |u(x)-u(y)| \leq 2\sqrt{r} \cdot\|u'\|_{L_2} \quad \forall x,y \in I. $$ The result follows by setting $r = 2|x-y|$ and using the fact that $C_c^1(\mathbb{R})$ is dense in $W^{1,2}(\mathbb{R})$.
What I belive is the definition of $\bar{u}$ should change depending on where you choose your interval $I$ to be centered, but in the argument above the author makes it seem that $\bar{u}$ does not change. If anyone could help me to understand the steps highlighted in red it would be very much appreciated.