The doubt arises from wanting to prove that $f(x)=\sqrt x$ is not Lipschitz continuous in $[0,1]$ by negating its definition: I know that proving this fact by contradiction is much simpler, but I would like to clarify this doubt regarding logic. Let's assume the negation of "$\exists L>0 \ \forall x,y \in [0,1],|f(x)-f(y)| \le L|x-y|$" is true, that is, assume: $$\forall L>0 \ \exists x_L,y_L \in[0,1], |f{x_L}-f{y_L}|>L|x_L-y_L|$$ To verify this proposition, I have thought of dividing it into two cases: $0<L\le 1/2$ or $L> 1/2$, and proceed by finding a pair $u_L, v_L \in [0,1]$ that satisfies the inequality $|f(u_L)-f(v_L)| > L|u_L-v_L|$ for $0 < L \le 1/2$, and another pair $r_L, s_L \in [0,1]$ that satisfies the inequality $|f(r_L)-f(s_L)| > L|r_L-s_L|$ for $L > 1/2$, and from this conclude that the negation is true for each $L>0$. However, upon reconsideration, I am not entirely convinced of the validity of this reasoning. My skepticism arises from the behavior of the universal quantifier with respect to logical disjunction: it's as if I am assuming that $\forall x \in X,(P(x)\vee Q(x)) \iff (\forall x\in X,P(x))\vee (\forall x\in X,Q(x))$.
However, when thinking of simple examples from everyday life, this logical equivalence seems false to me. For instance, if they were equivalent, the following statements would be equivalent: 'Every month of the year, there is a specific day when I eat fish' and 'In one of the months between January and June or in one of the months between July and December, there is a specific day when I eat fish.' The first statement implies the second one because since it occurs every month of the year, it also occurs specifically in one of the months between January and June or in one of the months between July and December. Therefore, the logical disjunction is true, and consequently, the second statement is true. However, if the second statement is true, then the logical disjunction is true, and it could happen that I exclusively eat fish in a month between January and June and then stop eating it from July to December, making the first statement false. That is, we only have $\forall x \in X,(P(x)\vee Q(x)) \implies (\forall x\in X,P(x))\vee (\forall x\in X,Q(x))$.
Hence, I cannot prove the Lipschitz continuity of $f$ dividing into in two cases $L>1/2$ or $0<L\le 1/2$ as I was thinking. Is my reasoning about the non equivalence of the logical statements correct or am I saying something wrong? Of course, any corrections, or suggestions, or simpler examples are welcome.
Copper.hat's comment is completely correct but I will still answer the question concerning the logical doubts.
You are right that $$ \forall x\in X, (P(x) \vee Q(x))$$ and $$ (\forall x\in X, P(x))\vee (\forall x\in X, Q(x))$$ are not equivalent.
For a mathematical example let $X\subset \mathbb{N}$ and define $P(x),Q(x)$ to be true if $x$ is even/odd and false if $x$ is odd/even respectively. Now the first statement holds for arbitrary $X$ as every natural number is odd or even. The second statement only holds if $X$ only contains numbers of one type. Hence the statements are not equivalent.
That being said, your proof would still be correct, as the logical structure is of another type. Let $X:=\{L>0\}$, $A:=\{L\leq 0,5\}$ and $B:=\{L>0,5\}$. Let $P(L)$ be the statement in your negation starting with $\exists x_L, y_L ...$. Now you show $\forall L\in A, P(L)$ and the same for $B$ and want to conclude $\forall L \in X, P(L)$. By the definition of a set-union we have $\forall L \in X, (L \in A) \vee (L \in B)$ hence it follows that $\forall L \in X, P(L)\vee P(L)$ and with that the claim you want to show.
Does that help?