Showing that $R(x)$ is a proper subset of $R((x))$ if $R$ is a field

HINT: Write $f_2(x)$ in the form $x^rg(x)$, where $g$ has a non-zero constant term. Then $g(x)$ has an inverse in $R[[x]]$.

An easy induction shows that its coefficients can be calculated recursively: just start calculating! For instance, if $g(x)=a_0+a_1x+\ldots+a_mx^m$, and the inverse is to be $h(x)=\sum_{k\ge 0}b_kx^k$, it’s clear that you want $b_0=a_0^{-1}$. Then the first degree term in $g(x)h(x)$ must be $$(a_0b_1+a_1b_0)x=(a_0b_1+a_0^{-1}a_1)x\;,$$

so $a_0b_1+a_0^{-1}a_1=0$, and you can solve for $b_1$. It’s easy to prove that this can be continued recursively.

And from there you’re pretty much home free.

In case $R$ is finite or countable, the rational-function field is countable, while the Laurent-series field is uncountable.

To show that $R(x)$ is a proper subset of $R((x))$, we first need to ignore the "is". Using more precision, I'd prefer to say that $R(x)$ is canonically isomorphic to a proper subring of $R((X))$.

First part: subset

We have a canonical and straightforward map from the ring $R[x]$ of polynomial to the ring $R((X))$ fo formal Laurent series (this does not even require $R$ to be a field) and accordingly identify polynomials with their corresponding power series. To extend this map to $R(x)$ we need to find, for every non-zero polynomial $f\in R[x]$, a series $u\in R((x))$ such that $f\cdot u=1$. First consider the case that $f$ has constant term $1$. Then we can define $u_i$, $i\in\Bbb N$, recursively such that for all $n$ $$\tag1 f(x)\cdot \sum_{i=0}^nu_ix^i\in 1+x^nR[x]$$ Indeed, we can just let $u_0=1$ and then recursively let $u_n$ be $-1$ times the coefficient of $x^n$ in the polynomial $f(x)\cdot\sum_{i=0}^{n-1}u_ix^i$. We obtain a power series $u(x)$ with $f(x)u(x)=1$ as desired.

Now consider general $f\ne 0$. Then it can be written as $a\cdot x^k\cdot \hat f$ where $a\in R\setminus \{0\}$, $k\in \Bbb N_0$, $\hat f$ is a polynomial with constant term $1$. As just seen, there is a power series $\hat u$ with $\hat f\hat u=1$. Then $u:=a^{-1}x^{-k}\hat u$ is a Laurent series with $fu=1$, as desired. (Here is the only place where we use that $R$ is a field: We need to find $a^{-1}$).

Remark: Actually, it suffices to know that $R((x))$ is itself a field; which by itself can be proved by finding a multiplicative inverse recursively (almost) precisely as above.

Second part: proper

It suffices to exhibit a single formal Laurent series that cannot be written as quotient of polynomials. Consider $$ u(x)=\sum_{k=0}^\infty x^{k^2} $$ and assume that $u=\frac fg$ with $g\ne 0$, say $g(x)=\sum_{j=0}^d a_jx^j$ with $a_d\ne 0$. Pick $m\ge \max\{d,1\}$. Then in multiplying $u(x)g(x)$ we see that the coefficient of $x^{m^2+d}$ equals $a_d$ because $\deg(x^kg)<m^2+d$ for $k<m$ and $x^{m^2+d+1}\mid x^kg$ for $k>m$. Hence $ug$ has infinitely many nonzero coefficients and is not a polynomial.