According to wikipedia if we have $f:X \rightarrow Y$ which is $\alpha$-Holderian then for all $\beta < \alpha$ the function is also $\beta$-Holderian.
How do we prove this starting from the fact that we have $d_y(f(x),f(y)) \leq K d_x(x,y)^{\alpha}$?
It seems we would need to show that $K d_x(x,y)^{\alpha}\leq K d_x(x,y)^{\beta}$ but $d_x(x,y) \leq 1$ does not always hold.
It depends on the definitions. If one wants a global Hölder constant, so that
$$d_Y(f(x),f(y)) \leqslant K\cdot d_X(x,y)^\beta$$
for all $x,y \in X$, then $\alpha$-Hölder continuity does in general not imply $\beta$-Hölder continuity for $\beta < \alpha$. The easy counterexample to that is $f(x) = \lvert x\rvert^\alpha$ on $\mathbb{R}$, which is $\alpha$-Hölder continuous, but doesn't admit a global Hölder constant for any exponent $\beta < \alpha$.
It is locally true, however, on every bounded subset $S$ of $X$ we have a $\beta$-Hölder constant when we have an $\alpha$-Hölder constant (the $\beta$-Hölder constant depends on the $\alpha$-Hölder constant and the diameter of $S$). Since Hölder continuity is mostly interesting for local behaviour, one can replace the metrics on $X$ and $Y$ by bounded uniformly equivalent metrics (for example $\bar{d}_X(x,y) = \min \{d_X(x,y),\, 1\}$) to obtain the implication.