I'm tring to solve this problem: prove that $$ u(x)=\left\{\begin{matrix} -\frac{1}{\log x}, & x>0\\ 0 & x=0 \end{matrix}\right. $$ is continuous but not in $\mathcal{C}^{0,\alpha}([0,1/2])$ for any $\alpha \in (0,1]$. My solution: $u$ is continuous on $[0,+\infty)$ as $\lim_{x \rightarrow 0^+} \frac{-1}{\log x}=0$. Regarding the second question, we have to prove that $$ [u]_{\mathcal{C}^{0,\alpha}([0,1/2])}= \sup_{x,y \in [0,\frac{1}{2}], \, x \neq y} \left \{ \frac{|u(x)-u(y)}{|x-y|^{\alpha}} \right \}= +\infty $$ Consider $$ \frac{|u(x)-u(y)|}{|x-y|^{\alpha}}=\frac{\left | -\frac{1}{\log x} +\frac{1}{\log y} \right |}{|x-y|^{\alpha}} \geqslant \left ( y+\frac{1}{\log x} \right ) \frac{1}{|x-y|^{\alpha}} $$ then taking the supremum of both sides we should have that $[u]_{\mathcal{C}^{0,\alpha}}=+\infty$ because $\frac{1}{|x-y|^{\alpha}}$ gets big as we want as $x \rightarrow y$.
My question is if my approach is correct, then I would like to see a different solution of this problem.
Your answer is not correct, as the inequality,
$$ \frac{\left| -\frac1{\log x} + \frac1{\log y}\right|}{|x-y|^{\alpha}} \leq \frac{\left(y+\frac1{\log x}\right)}{|x-y|^{\alpha}},$$
is not true in general. Indeed multiplying both sides by $|x-y|^{\alpha}$ gives,
$$ \left| -\frac1{\log x} + \frac1{\log y}\right| \leq \left(y+\frac1{\log x}\right).$$
This would imply however that $u$ is not continuous at any $x,$ which is evidently false.
You also didn't really prove that $u$ is continuous at 0, rather you just stated that it was. Admittedly this is very elementary so there's not much to say, but it's probably worth mentioning that $\lim_{x \rightarrow 0^+} \log x = -\infty.$
A general result states that if $f$ is continuously differentiable on a compact interval $I,$ then it is Hölder continuous there for all $\alpha \in [0,1].$ This follows by the mean value theorem, by noting noting $|f'|$ attains a maximum on $I$ and using it to deduce Hölder continuity.
If we look at our function $u,$ we have $u'(x) = \frac1{x (\log x)^2}$ on $(0,1/2].$ This is continuous, but standard calculations gives $u'(x) \rightarrow +\infty$ as $x \rightarrow 0^+.$ So we expect something to go wrong at $0.$ In fact, it will turn out that,
$$ \sup_{y \in (0,1/2]} \frac{|u(0)-u(y)|}{|y|^{\alpha}} = \infty, $$ for all $\alpha \in (0,1].$
To show this, the mean value theorem will be useful. Note that for $y \in (0,1/2],$ there is $\xi_y \in (0,y)$ such that,
$$ \sup_{z \in (0,1/2]} \frac{|u(0)-u(z)|}{|z|^{\alpha}} \geq \frac{|u(0)-u(y)|}{|y|^{\alpha}} = yu'(\xi_y) = \frac y{\xi_y^{1+\alpha} (\log \xi_y)^2} \geq \frac1{y^{\alpha} (\log y)^2}. $$
Where the last inequality holds since $u'$ is strictly decreasing on $(0,1/2).$ But this blows up at $y \rightarrow 0^+$ (again needs justifying), so we get Hölder continuity fails here.
Now I used two assertions here, namely that $u'$ is strictly decreasing on $(0,1/2)$ and that for $\alpha \in (0,1],$
$$ \lim_{x \rightarrow 0^+} \frac1{x^{\alpha} (\log x)^2} = \infty. $$
For the first we can compute $u''$ and show it's everywhere negative on $(0,1/2).$ For the second, it suffices to show that,
$$ \lim_{x \rightarrow 0^+} x^{\alpha} (\log x)^2 = 0,$$
which is a non-trivial computation left as an exercise for the reader. :) (Leave a comment if you get stuck and I'll provide an answer).