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: 
Manos, I know the frustration, so let me make another attempt in a different direction. I will do a simple case, where $n=3,s=2$. So, we have $V(F\_1,F\_2)=X\subset\mathbb{P}^3$, a smooth, irreducible complete intersection curve. Let $I\subset\mathcal{O}\_{\mathbb{P}^3}$ be its ideal sheaf. Then, we have the Koszul resolution, $0\to \mathcal{O}\_{\mathbb{P}^3}(-\deg F\_1-\deg F\_2)\to \mathcal{O}\_{\mathbb{P}^3}(-\deg F\_1)\oplus \mathcal{O}\_{\mathbb{P}^3}(-deg F\_2)\to I\to 0$, which by using the the fact that $H^i(\mathcal{O}\_{\mathbb{P}^3}(l))=0$ for $i=1,2$ and all $l$ gives $H^1(I(l))=0$ for all $l$ and the map $H^0(\mathcal{O}\_{\mathbb{P}^3}(-\deg F\_1+l)\oplus \mathcal{O}\_{\mathbb{P}^3}(-\deg F\_2+l))\to I(l)$ is onto for all $l$. Thus, we see that $\oplus I(l)\subset\oplus\mathcal{O}\_{\mathbb{P}^3}(l)=k[x\_0,x\_1,x\_2,x\_3]$ is just the ideal $(F\_1,F\_2)$. From the long exact sequence coming from $0\to I\to\mathcal{O}\_{\mathbb{P}^3}\to\mathcal{O}\_X\to 0$, using the vanishing of $H^1$, we get an exact sequence, $0\to (F\_1,F\_2)\to k[x\_0,x\_1,x\_2,x\_3]\to \oplus H^0(\mathcal{O}\_X(l))\to 0$. Using the fact that $X$ is integral, easy to check that so is $\oplus H^0(\mathcal{O}\_X(l))$, but this is precisely $k[x\_0,x\_1,x\_2,x\_3]/(F\_1,F\_2)$.
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.