You omitted one hypothesis: you’re trying to prove that if $\lambda\le\kappa$ and $\Bbb I$ is a $\kappa$-complete ideal on $\kappa$ containing all of the singletons, and if there is no pairwise disjoint family $\{X_\alpha:\alpha<\lambda\}\subseteq\wp(\kappa)\setminus\Bbb I$, then $\Bbb I$ is $\lambda$-saturated.
Suppose, on the contrary, that there is a family $\{X_\alpha:\alpha<\lambda\}\subseteq\wp(\kappa)\setminus\Bbb I$ such that $X_\alpha\cap X_\beta\in\Bbb I$ whenever $\alpha<\beta<\lambda$. Suppose first that $\lambda<\kappa$. Let $I=\bigcup\{X_\alpha\cap X_\beta:\alpha<\beta<\lambda\}$; $I$ is the union of $\lambda$ elements of $\Bbb I$, so $I\in\Bbb I$. For $\alpha<\lambda$ let $Y_\alpha=X_\alpha\setminus I$; clearly $Y_\alpha\notin\Bbb I$, and $Y_\alpha\cap Y_\beta=0$ if $\alpha<\beta<\lambda$. By hypothesis no such family $\{Y_\alpha:\alpha<\lambda\}$ exists, so we must have $\lambda=\kappa$.
For this case we need a slightly more sophisticated version of the same basic idea. For each $\alpha<\kappa$ let $Y_\alpha=X_\alpha\setminus\bigcup\{X_\alpha\cap X_\beta:\beta<\alpha\}\;;$ $|\alpha|<\kappa$, so $\bigcup\{X_\alpha\cap X_\beta:\beta<\alpha\}\in\Bbb I$, and therefore $Y_\alpha\notin\Bbb I$. But clearly $Y_\alpha\cap Y_\beta=0$ if $\alpha<\beta<\kappa$, since $Y_\alpha\cap Y_\beta\subseteq X_\alpha\cap \left(X_\beta\setminus\big(X_\beta\cap X_\alpha\big)\right)$, and again we contradict the hypothesis that no such family exists. This complete the proof.
Note that I could actually have done it in a single case using the more sophisticated version even when $\lambda<\kappa$; I just thought that it would be a little clearer if I started with the simple version of the idea.