Keep adding elements to $A$ until it's not possible anymore. Say that $A_i$ blocks an element $j$ if adding $j$ to $A$ would include $A_i$ in $A$. Clearly, this happens only if $|A \cap A_i| = 2$. This means at most one $A_i$ can block $j$ because any two such sets intersect in at most one element. So every blocked element can be mapped to a pair of elements in $A$. Since there are $n - |A|$ blocked elements, we have that $$n-|A| \leqslant \binom{|A|}{2}$$which gives that $|A| \geqslant \lfloor \sqrt{2n} \rfloor$, as desired. $\blacksquare$

This is just a generalized version of APMO 2008/2, the solution from the APMO problem works here as well.

I have a completely different idea, can someone check this works please? We will rewrite the problem in more geometrical way to understand the problem better. Let $X$ be a polygon with $n\ge 3$ vertices and let $A_1,A_2, \dots A_p$ be triangles by joining vertices of polygon such that no two triangle have no common sides. Show that there exists a polygon $A$ by joining vertices of polygon such that there is no any triangle $A_i$ "inside" it and number of vertices of $A$, which is $|A|$ holds $|A| \ge \sqrt[2]{2n}$ It is clear that $|A|$ is minimized when number of triangles such that no two triangle have no common vertices is maximized.Call these triangles $non-intersecting$ triangles, otherwise call them $intersecting$ triangles. Now, we will divide into cases. Case 1. $3|n$ Let $n=3k$. Hence, we can make total $k$ non-intersecting triangles, triangles, then $|A|= n-k=2k$, which is enough to show that $$ 2k \ge \lfloor{\sqrt{6k}} \rfloor$$Which is clearly true and equality holds when $k=1$. $\square$ Case 2. $n=3k+1$ Divide vertices into $2$ groups such that they have $3k$, $1$ elements. From $3k$ vertices, we can make $k$ non-intersecting triangles. Pair these triangles into total $ \lfloor \frac{k}{2} \rfloor $ pairs. For each pair, we can make $3$ intersecting triangles. Hence, there are total $k+ 3 \lfloor \frac{k}{2} \rfloor $ triangles. If $k=1$, there are no intersecting triangles and we can just check it manually, hence assume $k>1$. In order to remove intersecting triangles from our polygon's "inside", we must remove that $1$ extra point moreover , since we have $k$ non intersecting triangles, we must remove $k$ more vertices. Hence $$ n-k-1=2k \ge \lfloor \sqrt{6k+2} \rfloor $$Which is indeed true. $\square$ Case 3. $n=3k+2$. This follows similiarly as Case 2.. However, this time we have $2$ extra points. Similiar work shows that our inequality is indeed true.$\square$ Hence we are done!