Arbitrarily large finite models imply an infinite model

Add distinct new constant symbols $c_n$ for $n<\omega$ and put $S=T\cup \{c_i\mathrm{\ne }{c}_j:i<j<\omega \}$. This language remains explicitly countable by tagging old symbol codes and the indices of the new constants. A finite subset of $S$ mentions only finitely many new constants, say $m$ distinct ones. Take a finite model of $T$ of size at least $\max (1,m)$, interpret these constants by $m$ distinct elements, and interpret unused constants at one fixed element. The resulting expansion satisfies the fragment. This uses only one finite model and finitely many choices for this fragment.

F1 gives a model $N\models S$. Its interpretation map $n\mapsto c_n^N$ is injective, because every inequality is in $S$. Thus its carrier is infinite. The reduct to the original language still satisfies every sentence of $T$, and has the same carrier, so it is the required infinite model.