Question: Suppose I have a function that is continuous in every point of its domain, but its domain is only a subset of the real numbers (for example $\mathbb{R}-\{1\}$). Can I say that the function is continuous in $\mathbb{R}$? If not, what's the best way to express it?

Answer: 
First of all we can take $s \le n$. Since $F\_1,\dots,F\_s$ is a regular sequence, we have that $\operatorname{height}(I)=s$, where $I=(F\_1,\dots,F\_s)$, and also $R=S/I$ is Cohen-Macaulay.

Let $J$ be the Jacobian ideal of $F\_1,\dots,F\_s$, i.e. the ideal generated by all $s \times s$ minors of the Jacobian matrix of the $F\_i$. If $X = Z(F\_1,\dots,F\_s)$ is smooth, then $J \not\subset \mathfrak{m}\_x, \, \forall x \in X$ and so in the quotient ring $R$ we must have $\operatorname{height}(\bar{J}) \ge 1$. Finally, according to Theorem 18.15a in Eisenbud (CA, 2004) $I$ is reduced if and only if $\operatorname{height}(\bar{J}) \ge 1$.
If $F\_1,\ldots,F\_s$ are general, by Bertini theorem, we get that $\dim X=n-s$ and it is smooth. Thus, $k[x\_0,\ldots,x\_n]/(F\_1,\ldots,F\_s)$ is a complete intersection (follows from $\dim X=n-s$) and outside the irrelevant maximal ideal, this ring is regular. Since complete intersection implies the ring is Cohen-Macaulay, to see whether it is reduced, it suffices to do so at the minimal primes and this follows from the regularity at all the relevant primes.