Let $Y$ be a maximal subset of $X$ in the sense that if one adjoins another element from $X$ in $Y$ then it will contain a subset from $\mathcal{F}$. Define $f: X \setminus Y \to \binom{Y}{2}$, where $\binom{Y}{2}$ is the family of 2-subsets from $Y$, with $f(x) = A$ if $A \cup \{x\}$ is one of the sets from $\mathcal{F}$; if there is more than one set, choose any of them. $f$ is an injective function, because if $f(x_1) = f(x_2) = B$ then $B \cup \{x_1\}$ and $B \cup \{x_2\}$ would be in $\mathcal{F}$ and these two sets would have more than one element in their intersection. Thus, by the injective principle,
$$|X \setminus Y| \le \binom{|Y|}{2} \iff n - |Y| \le \frac{|Y|(|Y| - 1)}{2} \iff |Y| \ge -\frac{1}{2} + \sqrt{2n + \frac{1}{4}}$$
Let $\lfloor\sqrt{2n}\rfloor = k$. We need to prove that $-\frac{1}{2} + \sqrt{2n + \frac{1}{4}} > k - 1$ in order to prove that $|Y| > k - 1 \iff |Y| \ge k$. But this is only a small computation:
$$-\frac{1}{2} + \sqrt{2n + \frac{1}{4}} > k - 1 \iff 2n + \frac{1}{4} > k^2 - k + \frac{1}{4} \iff \sqrt{2n} > \sqrt{k^2 - k},$$
which is true because $\sqrt{2n} \ge k > \sqrt{k^2 - k}$.