In Royden's book there is:
It is easy to see that a sequence {$x_n$} in X converges to x $\in$ X with respect to the $\mathcal{F}$-weak topology iif $\lim \limits_{n \to \infty} f(x_n)=f(x)$ for all $f\in\mathcal{F}$
Here X is any set, $\mathcal{F}$ contains any function from X to R.
Logically this should be the quivalence to when $\mathcal{F}$ only contains one element. But for this one element situation, there need be an extra requirement that X is first countable for the reverse statement to hold.
Although any normed linear space as a metric space is first countable which means that the $\mathcal{F}$-weak topology (also $\mathcal{F}$'s elements are from X*) as a coarser topology than the norm-induced topology should be first countable too, which means the above quote is right for weak topology and weak-*topology, but in context of the above quote, X is only a set endowed with the $\mathcal{F}$-weak topology.
Does the book indeed lack the first countable property of the $\mathcal{F}$-weak topology there, or the first countable property can be induced from somewhere like the $\mathcal{F}$-weak topology itself (possibly from some properties of R, note that X is just a set)?
For the statement above we don't need first countability. I don't know if the weak topology in general is first countable or not.
$X$ is just a set, but the $\cal{F}$-weak topology equips it with a topology. At each $x\in X$, there is a base for this topology of the form $$\{x: |f_i(x)-f_i(y)|<\varepsilon\}$$ for some $\varepsilon>0$, $y\in X$ and some finite set $f_1,\dots, f_n$ in $\cal{F}$.
Assume $f(x_n)\to f(x_0)$ for every $f\in\cal{F}$. To prove that $x_n\to x_0$ we want to show that given an open set $U$ containing $x_0$, there exists $N$ such that if $n>N$,then $x_n\in U$. Since the above sets are a local base, there exist some $y\in X$,some $\varepsilon>0$ and some finite subset $\{f_1,\dots, f_m\}$ of $X$ such that if $|f_i(x_0)-f_i(y)|<\varepsilon$ for each $1\leq i\leq m$ then $y\in U$. By hypothesis, $f_i(x_n)$ converges to $f_i(x_0)$, for every $1\leq i\leq m$, so we can find $N$ such that if $n>N$, then $|f_i(x_n)-f_i(x_0)|<\varepsilon$, for all $1\leq i\leq m$, so $x_n\in U$.
For the other direction -- if $x_n\to x_0$ in the $\cal{F}$-weak topology, then for every open set $U$ containing $x_0$ there exists $N$ such that if $n>N$ then $x_n\in U$. Given $f\in\cal{F}$, and $\varepsilon>0$, the set $$\{x: |f(x)-f(x_0)|<\varepsilon\}$$ is an open set containing $x_0$, hence contains $x_n$ for sufficiently large $N$, which proves that $f(x_n)\to f(x_0)$.