Note that if $k_1\geq k$, then $f_{10}(j,k)\geq f_{10}(j,k_1)$. We have \begin{align*} f_{10}(n,n)&=f_{10}(n,n-1)\\ &\leq f_{10}(n,n-2)\\ &\leq f_{10}(n-1,n-2)\\ &\leq f_{10}(n-1,n-3)\\ &\leq f_{10}(n-2,n-3)\\ &\vdots\\ &\leq f_{10}(11,10)\end{align*}so $f_{10}(n,n)\leq f_{10}(j,j-1)$ for $11\leq j\leq n$. Suppose $f_{10}(j,n)<\frac12f_{10}(n,n)$ for some $1\leq j\leq n-1$. Then, there exists $1\leq i_1<i_2<\cdots<i_{11}\leq n$, with one equal to $j$ such that $A_jA_{i_m}<\frac12f_{10}(n,n)$ for all $1\leq m\leq11$, and $i_{11}\geq11$. Then, for $1\leq m\leq11$, we have $A_{i_{11}}A_{i_m}\leq A_{i_{11}}A_j+A_jA_{i_m}<f_{10}(n,n)$, so $A_{i_{11}}A_{i_1}$, $A_{i_{11}}A_{i_2}$, $\ldots$, $A_{i_{11}}A_{i_{10}}<f_{10}(n,n)\leq f_{10}(i_{11},i_{11}-1)$, contradiction.

Let $i\le n-1$, and let $j$ be one of the indices such that $d(i,j)=f(i,n)$. Then we have nine other points $k_1,k_2,\dots,k_9$ such that $d(i,k_\bullet)\le d(i,j)$. Assume WLOG $k_1<k_2<\dots<k_9$ and let $k=k_9$. Then note that $f(k,k-1)\le 2d(i,j)$. So \[f(i,n)=d(i,j)\ge\frac12 f(k,k-1)\ge\frac12 f(n,k-1)\ge\frac12 f(n,n)\;\blacksquare\]

Let $G$ be a graph on the vertices $A_i$ such that there is an edge between $A_i$ and $A_j$ when $A_iA_j<f_{10}(n,n)$. Claim: Each vertex $A_i$ has at most $9$ edges with $A_1,A_2,\dots A_{i-1}$. By problem statement, $$f_{10}(i,i-1)\geq f_{10}(n,i-1)\geq f_{10}(n,n)$$ since increasing the vertices from $i-1$ to $n$ clearly does not increase the $10$th smallest distance. However, this means that the $10$th closest to vertex $i$ within the first $i-1$ is already at least $f_{10}(n,n)$, so at most $9$ of them can have a distance less than $f_{10}(n,n)$, as desired. Now, suppose FTSOC there is a circle centered at $A_j$ with radius $\frac{1}{2}f_{10}(n,n)$ containing $11$ interior points (the center itself and its $10$ closest neighbors). Then, any two of these $11$ points are a distance of less than $f_{10}(n,n)$, so they form a $K_{11}$ in the graph $G$. However, we have shown that a vertex connects to at most $9$ previous vertices. This is a contradiction by taking the highest label in the $K_{11}$, as it would connect to $10$ vertices with smaller labels.

Fix $1 \le a \le n$. Draw a circle centered at $A_a$ with radius $f_{10}(a, n)$, and call the closed disk defined by such a circle $\mathbb D$. Note by definition there exist 11 points inside of $\mathbb D$, i.e. $1 \le i_1 < i_2 < \cdots < i_{11} \le n$ so that \[ A_{i_k} \in \mathbb D \quad \forall 1 \le k \le 11. \]In particular, there are at least $10$ indices $k$ less than $i_{11}$ so that $A_{i_{11}}A_{k} \le \operatorname{diam}(\mathbb D)$, and since $i_{11} \ge 11$ we have \[ 2f_{10}(a, n) = \operatorname{diam}(\mathbb D) \ge f_{10}(i_{11}, i_{11} - 1) \ge f_{10}(n, i_{11} - 1) \ge f_{10}(n, n), \]as desired.

Let $f_{10}(j,n)=A_jA_i$. This means that $f_{10}(j,k)=A_jA_i$ for all $i\le k\le n$. Also note that $f_{10}(j,a)\ge f_{10}(j,b)$ for all $a\le b$ so \[f_{10}(j,j-1)\ge f_{10}(n,j-1)\ge f_{10}(n,n)\]For all $j\ge 11$. In particular, if $i\le j-1$ we get \[f_{10}(j,i)\ge f_{10}(j,j-1)\ge f_{10}(n,n)\]Otherwise, there exist indices $i_1<i_2<\dots<i_9$ such that $A_jA_{i_1}<A_jA_{i_2}<\dots< A_jA_{i_9}<A_jA_i$. If $i>i_9$, then $A_{i_k}A_i\le A_{i_k}A_j+A_jA_i\le 2f_{10}(j,n)$, which implies $2f_{10}(j,n)\ge f(i,i-1)\ge f(n,n)$. The case $i<i_9$ can be solved similarly, hence we are done.

We fix $j$ and induct on $n$, with base case $n=\max(j,11)$. Base case: If $j>10$, then it remains to prove $f_{10}(j,j) \ge \tfrac{f_{10}(j,j)}{2}$, which is trivial. If $j \le 10$, then $n=11$, which means $f_{10}(k,n) \ge A_kA_m$ for any $k$ and $m$. If $f_{10}(n,n)=A_nA_k$, then we have \[f_{10}(n,n) \le A_jA_n+A_jA_k \le 2f_{10}(j,n),\]completing the base case. Inductive step: We prove $f_{10}(j,n) \ge \tfrac{f_{10}(n,n)}{2}$ given $j<n$ and $f_{10}(j,n-1) \ge \tfrac{f_{10}(n-1,n-1)}{2}$. We have two cases. $f_{10}(j,n)<A_jA_n$. Then, we have \[f_{10}(j,n)=f_{10}(j,n-1) \ge \frac{f(n-1,n-1)}{2} \ge \frac{f(n,n-1)}{2}=\frac{f(n,n)}{2},\]as desired. $f_{10}(j,n) \ge A_jA_n$. Then, let $k_0=n$, $k_1$, $\dots$, $k_9$ be the distinct indices other than $j$ with $f_{10}(j,n) \ge A_jA_{k_i}$ for $i=0,1,\dots,9$. Notice that $2f_{10}(j,n) \ge A_nA_j$ and \[2f_{10}(j,n) \ge A_jA_n+A_jA_{k_i} \ge A_nA_{k_i}\]for $i=1,2,\dots,9$. Since $n$, $j$, $k_1$, $\dots$, $k_9$ are distinct, we have $f_{10}(n,n) \le \max(A_nA_j,A_nA_{k_1},\dots,A_nA_{k_9})$. Therefore, \[2f_{10}(j,n) \ge \max(A_nA_j,A_nA_{k_1},\dots,A_nA_{k_9}) \ge f_{10}(n,n),\]as desired. We have completed the inductive step, so we are done. $\blacksquare$