Where you went wrong has been answered in a comment.
The result is not that hard. Your proof that a sequence in the unit ball of $C^{0,\alpha}$ must have a uniformly convergent subsequence is correct. So we may as well assume that $||f_n||_\alpha\le1$ and $f_n\to0$ uniformly, and we need to show that $||f_n||_\beta\to0$.
Let $\epsilon>0$. For every $n$ and every $x,y$ we have
$$|f_n(x)-f_n(y)|\le|x-y|^\alpha=|x-y|^{\alpha-\beta}|x-y|^\beta.$$Choose $\delta>0$ so that $\delta^{\alpha-\beta}=\epsilon$. Then the above shows that $$|f_n(x)-f_n(y)|\le\epsilon|x-y|^\beta\quad(|x-y|\le\delta).$$On the other hand, if $n$ is large enough then $2||f_n||_\infty/\delta^\beta<\epsilon$, so that $$|f_n(x)-f_n(y)|\le2||f_n||_\infty\le\frac{2||f_n||_\infty}{\delta^\beta}|x-y|^\beta<\epsilon|x-y|^\beta\quad(|x-y|>\delta).$$
(Similarly if $f_n\to g$ uniformly...)
Lemma: Let $X,Y$ be Banach spaces. If $T_n:X\to Y$ are bounded operators with finite dimensional range and $T_n\to T\in B(X,Y)$ in operator norm, then $T$ is a compact operator.
For the case $X=Y=L^p$, $1< p\le\infty$, we will write $T$ as a norm limit of finite rank operators and deduce compactness of $T$ by the above lemma.
Let $n>1$ be an integer and partition the interval $[0,1]$ in $n$ subintervals of equal length, namely $I_{j,n}:=[\frac{j-1}{n},\frac{j}{n})$, $j=1,\dots,n$. Let $\phi_{j,n}$ denote the indicator function of $I_{j,n}$. Set
$$T_n:L^p[0,1]\to L^p[0,1], \;\;\;T_n(f)=\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}$$
Note that the operator $T_n$ has automatically finite dimensional range, since $T_n(f)$ lies in the linear span of the functions $\{\phi_{j,n}:j=1,\dots,n\}\subset L^p[0,1]$. Also,
for $x\in[0,1]$ there exists a unique $j_x\in\{1,\dots,n\}$ such that $x\in I_{j_x,n}$ so $\phi_{i,n}(x)=1$ if and only if $i=j_x$. So
$$|T_n(f)(x)|^p=\bigg|\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}(x)\bigg|^p=\bigg|\int_0^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg(\int_0^1|f(t)|dt\bigg)^p\le\|f\|_p^p$$
where in the last inequality we have used the fact that $\|f\|_1\le\|f\|_p$ for $f\in L^p[0,1]$. Thus
$$\|T_n(f)\|_p^p=\int_0^1|T_n(f)(x)|^pdx\le\int_0^1\|f\|_p^pdx=\|f\|_p^p$$
and thus $\|T_n\|\le1$, so $T_n$ are indeed bounded operators with finite rank.
We now show that $\|T_n-T\|_p\to0$. Indeed, we have
$$|T_n(f)(x)-T(f)(x)|^p=\bigg|\sum_{j=1}^n\int_0^{\frac{j}{n}}f(t)dt\cdot \phi_{j,n}(x)-\int_0^xf(t)dt\bigg|^p=$$ $$=\bigg|\int_0^\frac{j_x}{n}f(t)dt-\int_0^xf(t)dt\bigg|^p\;\;(\star)$$
where $j_x\in\{1,\dots,n\}$ is the unique integer such that $x\in I_{j_x,n}$ (and the above equality occurs because $\phi_{j_x,n}(x)=1$ and $\phi_{i,n}(x)=0$ for $i\ne j_x$). Continuing from $(\star)$, if we denote by $q$ the conjugate exponent $(1/p+1/q=1)$, we have
$$(\star)=\bigg|\int_x^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg|\int_0^1\chi_{I_{j_x,n}}(t)f(t)dt\bigg|^p\le$$ $$\le\bigg(\int_0^1\chi_{I_{j_x,n}}(t)\cdot|f(t)|dt\bigg)^p\le\bigg(\mu(I_{j_x,n})^{1/q}\cdot\|f\|_p\bigg)^p=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$
where we used Holder's inequality. Therefore,
$$\|T_n(f)-T(f)\|_p^p=\int_0^1|T_n(f)(x)-T(f)(x)|^pdx\le\int_0^1\frac{1}{n^{p/q}}\cdot\|f\|_p^pdt=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$
and thus $\|T_n-T\|_p\le\frac{1}{n^{1/q}}$. Letting $n\to\infty$ gives $T_n\to T$.
P.S:
Why it is reasonable to define the operators $T_n$ the way we did? First, we need them to have finite dimensional range. Second, we look at $T(f)(x)=\int_0^xf(t)dt$. This is a number very close to $\int_0^{j/n}f(t)dt$ for some suitable $j,n$. So it feels natural to partition the unit interval in small intervals of length $1/n$ and define $T_n(f)$ by the rule "take a $x$, determine in which small interval it lies (i.e. find the proper $j_x$), then assign the value $\int_0^{j_x/n}f(t)dx$. Implicitly we have been multiplying with $\phi_{j,n}$ and adding up, to make sure we evetually obtained the correct $j_x$. I hope this helps you understand the reasoning here.
Best Answer
The integral from $a$ to $b$ can be viewed as $\int_a^bf=\int_{0}^1\chi_{[a,b]}f$. Recall the relation $$ |\chi_A-\chi_B| = \chi_{A \Delta B} $$ where $A\Delta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that $$ [a,b] \Delta [c,d] \subset [a,c] \cup[c,a]\cup[b,d]\cup[d,b] $$ where $[x,y]=\emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $a\le c$ and $b\le d$.
Using the above, we can compute that $$\begin{align} \left|\int_a^bf-\int_c^df \right| &\le \int_0^1 |\chi_{[a,b]} - \chi_{[c,d]}| \,|f|\\ &= \int_0^1 \chi_{[a,b] \Delta [c,d]} |f|\\ &\le\int_0^1 \chi_{[a,c]} |f| + \int_0^1 \chi_{[b,d]}|f| \\ &\le (|c-a|+|d-b|)\sup_{x\in[0,1]} |f(x)|. \end{align}$$
Back to your question, we can deduce that $$\begin{align} |Tu(x)-Tu(y)| &\le \big(|\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)| \big)\, ||u||_\infty \\ &\le |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)| \end{align}$$ for any $u\in B$. By continuity of $\alpha$ and $\beta$, for any given $\varepsilon>0$ we may find $\delta$ such that $$ |x-y|<\delta \implies |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)|<\varepsilon, $$ which proves what you want.