a net of quadrics and the corresponding intersection Let $Q_i(i=1,2,3)$ be quadric hypersurface in $\mathbb{P}^4$. Consider a net of quadrics
$\Lambda=(Q_1,Q_2,Q_3)$.
I can't understand some part of proof of Corollary 2.8(p.11) in Stability of genus 5 canonical curves.

Corollary 2.8  If a net of quadrics in $\mathbb{P}^4$ is semistable, then the corresponding intersection is connected and purely 1-dimensional. 

My question is:
(1) They said "Fulton-Hansen connectedness theorem gives the first statement." But to use the theorem, I need irreducibility of quadrics. How can I use the theorem? Can I choose a basis $\{Q_1,Q_2,Q_3\}$ of the net such that each $Q_i$ is irreducible? 
(2) Assume that we can choose a basis $\{Q_1,Q_2,Q_3\}$ of the net, such that $S:=Q_1\cap Q_2$ is a quartic surface and there is an irreducible component $S'$ of $S$ which is contained in $Q_3$. If $deg(S')=2$, then why the span of $S'$ is a hyperplane?
 A: (1) Yes.
Consider a net of reducible quadrics (here the dimension of the projective space is not relevant): since all the associate quadratic forms have rank at most 2, and since they vary linearly, then there is a vector subspace of codimension 2 contained in the kernel of all the quadratic forms in the net. Then, in suitable coordinates, it is the net $ \langle x_0^2, x_0x_1, x_1^2 \rangle$ which is not semistable.
(2) Every nondegenerate (=not contained in any hyperplane) variety in ${\mathbb P}^r$ has degree strictly bigger than its codimension (so a nondegenerate surface in  ${\mathbb P}^4$ has degree strictly at least 2+1=3). 
This can be easily proved by induction on the codimension: it is trivial in codimension 1, and the induction step can be easily performed by projecting the variety from a smooth point, reducing both the codimension and the degree by 1.
A: An ample divisor on a variety of dimension $\ge 2$ is always connected, see Cor. III.7.9 in Hartshorne. For the second see Ex. I.7.8.
