Let $F=f^{-1}(0)\subset \mathbb{R}^n$.
It is a closed subset, which disconnects $\mathbb{R}^n$, and all connected components of the complement are unbounded.
The question amounts to see if the connected component of $\infty$ in $\widehat{F}=F\cup\infty \subset S^n$ could be reduced to $\infty$.
If it were the case, there would be a decreasing sequence $V_i$ of neighbourhoods of $\infty$ with intersection only this point and $Fr(V_i)$ disjoint from $F$ (in a compact space, quasicomponents are the same as connected components).
But then $K_i=F\setminus V_i$$K_i=F- V_i$ is compact and does not disconnect $\mathbb{R}^n$, otherwise there would be a bounded component of the complement, since it would also disconnect $S^n$ ($n>1$ is used here). But this easily implies that $\mathbb{R}^n-F$ would have a bounded component, contrary to the assumptions.
Hence $F$ is the union of a sequence of disjoint non-disconnecting compact sets $L_i=K_i\setminus K_{i-1}$ going to infinity.
Now observe that $L_i$ doesn't disconnect any open connected subset $U$ containing it, by excision. More precisely this gives an isomorphism of relative homology groups $H_1(U_i,U_i-L_i)\simeq H_1(\mathbb{R}^n,\mathbb{R}^n-L_i)\simeq \widetilde{H}_0(\mathbb{R}^n-L_i)=0$$$H_1(U_i,U_i-L_i)\simeq H_1(\mathbb{R}^n,\mathbb{R}^n-L_i)\simeq \widetilde{H}_0(\mathbb{R}^n-L_i)=0$$, where the second isomorphism comes from the homology exact sequence of the pair $(\mathbb{R}^n,\mathbb{R}^n-L_i)$. Then the homology exact sequence for $(U_i,U_i-L_i)$ gives $\widetilde{H}_0(U_i-L_i)=0$ for a connected $U_i$, as claimed.
It remains to take disjoint connected $U_i$'s containing $L_i$'s, with union $U$ and then $H_1(U,U-F)\simeq \bigoplus_i H_1(U_i,U_i-L_i) =0$,$$H_1(U,U-F)\simeq \bigoplus_i H_1(U_i,U_i-L_i) =0,$$ a contradiction since this is also $H_1(\mathbb{R}^n,\mathbb{R}^n-F)\simeq \widetilde{H}_0(\mathbb{R}^n-F)\neq 0$.$$H_1(\mathbb{R}^n,\mathbb{R}^n-F)\simeq \widetilde{H}_0(\mathbb{R}^n-F)\neq 0\ .$$