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Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. Your chosen method of proof is a good one, and you are almost there - just one more step to go!

Theorem 1: If $F_X(z) > F_Y(z)$ for all $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: We will proceed using a proof-by-contradiction. Contrary to the result in the theorem, suppose that $\mathbb{P}(X<Y)=0$. Then for any $z \in \mathbb{R}$ you have: $$\begin{equation} \begin{aligned} F_X(z) = \mathbb{P}(X \leqslant z) &= \mathbb{P}(X \leqslant z, X < Y) + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= 0 + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &\leqslant \mathbb{P}(Y \leqslant z) = F_Y(z), \\[6pt] \end{aligned} \end{equation}$$ which contradicts the antecedent condition for the theorem. $\blacksquare$

It is worth noting here that we can prove a stronger result that this one. Since our proof by contradiction shows a contradiction to the antecedent condition for all $z \in \mathbb{R}$, we can weaken the antecedent condition to a weaker condition than even the standard condition for strict first-order stochastic dominance.

Theorem 2: If $F_X(z) > F_Y(z)$ for some $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: As above.

Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. This condition is much stronger than what you actually need to get the result in the conjecture, so I will give you a proof for a stronger result (same implication but with a weaker antecedent condition). Your chosen method of proof is a good one, and you are almost there - just one more step to go!

Theorem: If $F_X(z) > F_Y(z)$ for some $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: We will proceed using a proof-by-contradiction. Contrary to the result in the theorem, suppose that $\mathbb{P}(X<Y)=0$. Then for all $z \in \mathbb{R}$ you have: $$\begin{equation} \begin{aligned} F_X(z) = \mathbb{P}(X \leqslant z) &= \mathbb{P}(X \leqslant z, X < Y) + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= 0 + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &\leqslant \mathbb{P}(Y \leqslant z) = F_Y(z), \\[6pt] \end{aligned} \end{equation}$$ which contradicts the antecedent condition for the theorem. This establishes the theorem by contradiction. $\blacksquare$

Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. Your chosen method of proof is a good one, and you are almost there - just one more step to go!

Theorem: If $F_X(z) > F_Y(z)$ for all $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: We will proceed using a proof-by-contradction. Contrary to the result in the theorem, suppose that $\mathbb{P}(X<Y)=0$. Then for any $z \in \mathbb{R}$ you have: $$\begin{equation} \begin{aligned} F_X(z) = \mathbb{P}(X \leqslant z) &= \mathbb{P}(X \leqslant z, X < Y) + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= 0 + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &\leqslant \mathbb{P}(Y \leqslant z) = F_Y(z), \\[6pt] \end{aligned} \end{equation}$$ which contradicts the antecedent condition for the theorem. $\blacksquare$

It is worth noting here that we can prove a stronger result that this one. Since our proof by contradiction shows a contradiction to the antecedent condition for all $z \in \mathbb{R}$, we can weaken the antecedent condition to the standard condition for strict first-order stochastic dominance. Thus, the method also serves as a valid means of proving the following stronger theorem.

Theorem 2: If $X \ll Y$ (strict first-order stochastic dominance) then $\mathbb{P}(X<Y) > 0$.

Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship. Your chosen method of proof is a good one, and you are almost there - just one more step to go!

Theorem 1: If $F_X(z) > F_Y(z)$ for all $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: We will proceed using a proof-by-contradiction. Contrary to the result in the theorem, suppose that $\mathbb{P}(X<Y)=0$. Then for any $z \in \mathbb{R}$ you have: $$\begin{equation} \begin{aligned} F_X(z) = \mathbb{P}(X \leqslant z) &= \mathbb{P}(X \leqslant z, X < Y) + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= 0 + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &= \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt] &\leqslant \mathbb{P}(Y \leqslant z) = F_Y(z), \\[6pt] \end{aligned} \end{equation}$$ which contradicts the antecedent condition for the theorem. $\blacksquare$

It is worth noting here that we can prove a stronger result that this one. Since our proof by contradiction shows a contradiction to the antecedent condition for all $z \in \mathbb{R}$, we can weaken the antecedent condition to a weaker condition than even the standard condition for strict first-order stochastic dominance.

Theorem 2: If $F_X(z) > F_Y(z)$ for some $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.

Proof: As above.