= Solution
An <almost complex structure> is a smooth real <bundle endomorphism> $J:TM\to TM$ satisfying $J^2=-I$. In particular the real dimension is even, say $2m$. After complexifying, the <type decomposition of the complexified tangent bundle> is
$$
T_{\mathbb C}M=T^{1,0}M\oplus T^{0,1}M,
\qquad T^{1,0}M=\ker(J-i),\quad T^{0,1}M=\ker(J+i).
$$
The corresponding <differential forms of type (p, q)> are smooth sections of $\Lambda^p(T^{1,0}M)^*\otimes\Lambda^q(T^{0,1}M)^*$, regarded as alternating forms by the <wedge product of differential forms>. A form of type $(p,q)$ vanishes unless its arguments include $p$ vectors of type $(1,0)$ and $q$ of type $(0,1)$.
Besides vanishing of the <Nijenhuis tensor>, two equivalent descriptions of an <integrable almost complex structure> are: the smooth sections of $T^{0,1}M$ are closed under <Lie bracket>; and every point has complex coordinates in which $J$ is multiplication by $i$, with <holomorphic> changes of coordinates. The equivalence with existence of these coordinates is the <Newlander-Nirenberg theorem> for smooth $J$. A further equivalent condition is that the <exterior derivative> has only the two type components $d=\partial+\bar\partial$, with no components of bidegree $(2,-1)$ or $(-1,2)$. Indeed evaluation of $d$ of a $(1,0)$-form on two $(0,1)$ vector fields detects the $(1,0)$ component of their bracket; <complex conjugation> detects the other component.
For the <almost complex structure induced by a complex atlas>, write $z^j=x^j+iy^j$ and put $J\partial_{x^j}=\partial_{y^j}$, $J\partial_{y^j}=-\partial_{x^j}$. The differentials of <holomorphic> transition maps are complex linear, so these local definitions agree. In these coordinates $J$ has constant coefficients and all coordinate vector fields commute. The expression defining the <Nijenhuis tensor> is tensorial, so evaluating it on this frame proves it is zero everywhere. Thus the induced structure is integrable.
Here is a direct proof for the <smooth submanifolds with invariant complex tangent spaces are complex> assertion. At a point of $Y$, its <tangent space> is a complex-linear subspace, of real dimension $2k$. Make a complex-linear change of ambient <holomorphic coordinates> so that this <tangent space> is $\mathbb C^k\times0$. Projection onto the first $k$ coordinates has invertible real differential on $Y$. The real <inverse function theorem> states that a smooth map between equal-dimensional manifolds with invertible differential is a <diffeomorphism> on sufficiently small neighbourhoods. It therefore writes $Y$ locally as a smooth graph $w=F(z)$ over an open set in $\mathbb C^k$.
A tangent vector to this graph is $(v,dF(v))$. Its image under $J$ is $(iv,i\,dF(v))$. Tangent-space invariance implies
$$
dF(iv)=i\,dF(v),
$$
so $F$ satisfies the <Cauchy-Riemann equations>. Smooth functions satisfying these equations are <holomorphic>: restrict to each coordinate complex line, use the one-variable criterion, and apply the iterated Cauchy integral formula on a small polydisc to get jointly <holomorphic> coefficients. Hence the graph parametrization is <holomorphic>. Restrictions of ambient <holomorphic coordinates> supply <holomorphic> transition maps between these graph charts. \b[The restricted structure on $Y$ is an integrable <complex structure>.] This also proves the induced structure agrees with the original restriction, rather than merely producing some <complex structure> on $Y$.
A <nonsingular analytic subvariety> is a closed analytic subset which near each of its points is the zero set of a <holomorphic> submersion of constant rank; equivalently, it is locally a coordinate subspace in <holomorphic coordinates>. The <holomorphic implicit function theorem> says that if a <holomorphic map> to $\mathbb C^r$ has differential of rank $r$ at a zero, its nearby zero set is a complex submanifold of codimension $r$.
The homogeneous equation defines a subset of <Complex projective space> because multiplying a vector by $\lambda\ne0$ multiplies its quadratic value by $\lambda^2$. Its gradient on $\mathbb C^4$ is $2Qz$, which is nonzero for $z\ne0$. More explicitly, in an affine chart set $z_i=1$. If all derivatives of the restricted equation vanished at one of its zeros, all components of $2Qz$ except possibly the $i$th would vanish. The <Euler homogeneous function theorem> gives $\sum_j z_j\partial_jp=2p=0$, forcing the remaining component to vanish as well. This contradicts invertibility of $Q$. The <holomorphic implicit function theorem> thus proves \b[$S$ is a smooth complex surface].
To see its geometry, reduce the nondegenerate symmetric bilinear form to diagonal form over $\mathbb C$. There is a vector of nonzero self-pairing, since otherwise polarization would make the entire form zero. Normalize such a vector using a square root of its self-pairing. Its <orthogonal complement> remains nondegenerate, so induction gives an orthonormal basis. Apply the same argument to the nondegenerate form $w_0w_3-w_1w_2$: the two forms are equivalent under an invertible linear change of variables. The allowed <projective linear transformation> therefore reduces $S$ to
$$
Q_0=\{[w_0:w_1:w_2:w_3]:w_0w_3-w_1w_2=0\}.
$$
View its coordinates as the nonzero rank-one matrix $A=\begin{pmatrix}w_0&w_1\\w_2&w_3\end{pmatrix}=uv^T$. At $[A]$, holding $[u]$ fixed and varying $[v]$ gives a <projective line>, and holding $[v]$ fixed and varying $[u]$ gives another. They are distinct and meet at $[A]$. In fact these are the only lines through it: if a second rank-one matrix is $ab^T$, the entire pencil has determinant zero only if $\det(u,a)\det(v,b)=0$, so one projective factor is fixed. These are the two <rulings of a smooth quadric surface>.
The <Segre embedding> is the <holomorphic map>
$$
\Phi:\mathbb{CP}^1\times\mathbb{CP}^1\longrightarrow Q_0,
\qquad([u_0:u_1],[v_0:v_1])\longmapsto[u_0v_0:u_0v_1:u_1v_0:u_1v_1].
$$
Every nonzero rank-one matrix has such a factorization, unique up to reciprocal rescaling, so $\Phi$ is bijective. Its inverse is <holomorphic> on each open set where a matrix entry $a_{ij}\ne0$: the projective column $[a_{0j}:a_{1j}]$ recovers $[u]$, and the projective row $[a_{i0}:a_{i1}]$ recovers $[v]$. These formulas agree on overlaps. Thus the <Segre description of a smooth quadric surface> gives the promised stronger conclusion
$$
\boxed{S\cong\mathbb{CP}^1\times\mathbb{CP}^1\quad\text{biholomorphically}.}
$$
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