At the risk that I'm doing somebody's homework...

Yes, $R_{m,n}[a,b]$ is complete with respect to the uniform metric
$\left\| \cdot \right\|$.

Let $f \in {\cal C}([a,b])$ be a uniform limit of rational functions
$f_j = p_j \phantom./\phantom. q_j \in R_{m,n}$ with each $p_j$ (respectively $q_j$)
a polynomial of degree at most $m$ (resp. $n$), and $q_j(x) \neq 0$
for all $x \in [a,b]$. We shall show that $f \in R_{m,n}[a,b]$.

We would like to write $f = p/q$ with $p = \lim_j p_j$ and $q = \lim_j q_j$,
but this doesn't work in general because $f_j$ does not determine
$p_j,q_j$ uniquely: we can multiply $p_j,q_j$ by the same scalar,
and if $f_j$ is actually a quotient of polynomials of degrees
strictly less than $m$ and $n$ then we could even introduce a
nonconstant common factor into $p_j$ and $q_j$.

We deal with
the first ambiguity by scaling each $p_j$ and $q_j$ so that
$\left\| q_j \right\| = 1$. Then the $q_j$ are all on the
unit sphere in the finite-dimensional space $P_n[a,b]$,
which is compact. This lets us finesse the second ambiguity
by passing to a subsequence that converges in ${\cal C}([a,b])$.

Let $q$, then, be the limit of such a subsequence $\lbrace q_k \rbrace$,
and let $p = f q$. Then $p$ is the uniform limit of polynomials
$p_k = f_k q_k$, because $f$ and $q$ are uniform limits of $f_k$ and $q_k$
respectively. But each $p_k$ is in the finite-dimensional vector space
$P_m$, which is complete in ${\cal C}([a,b])$; so the uniform limit
$p$ is also a polynomial, of degree at most $m$.

Let $f_\infty \in R_{m,n}$ be the rational function $p/q$,
*reduced to lowest terms*. We claim that $f = f_\infty$ on $[a,b]$.
This is not quite immediate from what we've done so far, because
$q$ might have a zero $x_0 \in [a,b]$ even though none of the $q_k$ does.
[For example, let $m=n=2$ and $f_k = 1 + 1/(k^2 (x-x_0)^2 + k)$.]
Still, since $f_k = p_k/q_k$ with $p_k \rightarrow p$ and $q_k \rightarrow q$,
it does follow that $f_\infty(x) = f(x)$ for all $x \in [a,b]$
where $q(x) \neq 0$. Since $f$ is continuous on $[a,b]$, it follows
that $f_\infty(x) = f(x)$ even at each of the finitely many $x$ where
$q(x) = 0$. **QED**