Well, i am not sure if this is what the OP is looking for but here is an heuristic method for computing the limit, avoiding the use of another algebra defined at $q=1$ and thus bypassing the "double cover problem" (mentioned in the OP):
Let us use the standard presentations in terms of generators and relations,
- $U\big(sl(2)\big)$ is:
$\ \ \ [H,X]=2X$, $[H,Y]=-2Y$, $[X,Y]=H$
whereas
- $U_q\big(sl(2)\big)$ is:
$\ \ KK^{-1}=K^{-1}K=1$, $KE=q^2EK$, $KF=q^{-2}FK$, $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$
The problem is that we cannot just plug-in $q=1$ in the above relations because the limit is indeterminate. In order to work around this problem (essentially following Drinfeld's original approach), let us use the "change of variables":
$$
q=e^{h/2} \ \ \ \ K=q^H=e^{hH/2}
$$
These imply: $$\frac{dK}{dh}=\frac{HK}{2}, \ \ \ \ \ \ \ \ \ \ \lim_{h\to 0}K=1$$
Now
$$
[K,E]=KE-EK=(q^2-1)EK=(e^h-1)EK
$$
Differentiating the last relation wrt to $h$ gives:
$$
\frac{1}{2}[HK,E]=e^hEK+(e^h-1)E\frac{h}{2}K
$$
and finally take the limit of the above while $h\to 0$ to get:
$$[H,E]=2E$$
With the correspondence $E\leftrightarrow X$ and $F\leftrightarrow Y$ (the $[K,F]$ relation is treated in a similar manner) these give the first two relations of $U\big(sl(2)\big)$.
Now the third relation is written as
$$
[E,F]=\frac{K-K^{-1}}{q-q^{-1}}=\frac{e^{hH/2}-e^{-hH/2}}{e^{h/2}-e^{-h/2}}
$$
Take the limit of both sides at $h\to 0$, using Del' Hospital in the rhs:
$$
[E,F]=\lim_{h\to 0}\frac{\frac{HK}{2}-(-\frac{HK}{2})}{\frac{1}{2}e^{h/2}-(-\frac{1}{2}e^{-h/2})}=H
$$
which gives the third relation describing the multiplication of the $U\big(sl(2)\big)$ algebra as the $q\to 1$ limit of the multiplication of the $U_q\big(sl(2)\big)$ algebra.
Finally, the coalgebra structure (comultiplication, counity) and the antipode limits can be handled in a similar manner to conclude that the $q\to 1$ limit of the Quantum group $U_q\big(sl(2)\big)$ is the Hopf algebra $U\big(sl(2)\big)$.
Edit: Regarding your last question:
Is there some reason why people felt it best to stick with the double cover approach?
i would say that (contrary to the remark in the OP:
This fact goes on to create many problems, such as the effective doubling of the number of representations ...)
this "doubling" of the representations is the reason behind the double-cover approach in the sense that this provides a more direct analogy between the representations of $U_q\big(sl(2)\big)$ and the representations of $U\big(sl(2)\big)$: The representation theory of $U_q$ includes two classes of highest weight modules, parameterized by the positive integers and the sign $\epsilon =\pm 1$, unlike the rep theory of $U$ which includes a single class of highest weight modules parameterized by the positive integers.
(This happens genererally for $q$ being either a root of unity or not, although for the root of unity case things are a little more subtle i.e. there are other classes of reps as well, an upper bound in their dim etc; but this is irrelevant to the rest).
So, when taking the $q\to 1$ limit of $U_q$, its $\epsilon =1$ reps give the usual $U$-highest weight modules while its $\epsilon =-1$ reps correspond to the "double" class of reps of $U(\mathfrak{sl}_2) \otimes \mathbb{Z}/2$ (since the last algebra can be viewed as an extension of $U$ by adjoining a single, central generator satisfying $g^2=1$).
