Open. No source found proves or refutes eventual monotonicity of $f$. The size of $f$ is known: $f(n)=(1+o(1))\sqrt n$ (the site, from the bounds of Problem 552; Erdős's display (3) of 1996 for $h(n)$) and $f(4)=2$ (the site). The site's further bound $f(n)<\sqrt n+1$ is an off-by-one error: the counting argument behind it bounds $h$, the largest minimum degree of a $C_4$-free graph on $n$ vertices, by $h(n)(h(n)-1)\le n-1$, so $h(n)\le(1+\sqrt{4n-3})/2$ and $f(n)=h(n)+1<\sqrt n+3/2$, and Erdős's "$f(n)<\sqrt n+1$ is easy" in [Er93] is about his own $f$, which is $h$. The bound fails for the page's $f$: the line graph of the Petersen graph is $4$-regular and $C_4$-free on $15$ vertices, so $f(15)=5>\sqrt{15}+1$ (Boza's values $R(C_4,K_{1,10})=14$ and $R(C_4,K_{1,11})=16$ give the same through the corrected conversion formula). The closest statements concern the equivalent star Ramsey sequence $s(n)=R(C_4,K_{1,n})$: Boza's Remark 12 records $s(n)\ge s(n-1)+1$ for $3\le n\le39$ with no counterexample known for larger $n$, and Chen's Theorem 4 [Ch97] gives $s(n+1)\le s(n)+2$ for all positive integers $n$ (Boza's Lemma 1 quotes it as $s(n-1)\ge s(n)-2$). Since $s$ is nondecreasing, the corrected conversion gives $f(n)=n-\max\{k:s(k)\le n\}$ for $n\ge4$, so $f(n+1)<f(n)$ exactly when $s(k)=s(k+1)=n+1$ for some $k$. The problem is therefore equivalent to $s(k+1)>s(k)$ for all large $k$, the negation of the question of Burr, Erdős, Faudree, Rousseau and Schelp whether $s(k+1)=s(k)$ holds infinitely often ([BEFRS89], p. 89, which also asks whether such $k$ have density zero; the site records the question under Problem 552). Boza's Remark 12 is that inequality for $2\le k\le38$, which with $s(38)=45$ gives $f(n+1)\ge f(n)$ for $4\le n\le44$; Chen's theorem gives $f(n+2)\le f(n)+1$. The search, whose scope the Current assessment records, found nothing more. This is a bounded negative finding, not a certificate of openness.