Why a subvariety of a variety of general type is of general type

You need to be careful about what you mean by "a general point". Usually, this means "a point in a certain Zariski open set". So in particular, this statement would say that on a surface of general type all rational and elliptic curves lie in a Zariski closed subset. This is Lang's conjecture, still open I believe (proofs were suggested about 15 yrs ago but then withdrawn).

EDIT: OK, so from the comments and the other answer it appears that it should be a "very general point of $X$", and the statement reduces to showing that if you have a morphism $f:Y\to T$ with irreducible $T$ and general fiber $Y_t$, and a finite dominant morphism $\pi:Y\to X$, then $Y_t$ is also of general type.

The basic reason for that is very simple: if $X$ is of general type then it has lots of pluricanonical forms. You can pull them back to $Y$ (they are differential forms, after all) and get lots of pluricanonical forms on $Y$. Then you can restrict them to $Y_t$ and get lots of pluricaninical forms on $Y_t$.

For a more precise answer, I suggest you look at old papers by Kawamata, Viehweg and Kollar, search for "additivity of Kodaira dimension". There is a whole sequence of $C_{n,m}$ conjectures about the Kodaira dimension of a fibration $Y$ in terms of Kodaira dimensions of $T$ and $Y_t$. Some of them are proved, some are still open.

(Note: general type means "maximal Kodaira dimension", i.e. equal to the dimension of the variety.)

Let me reproduce the relevant bit from the reference above (Internet Archive):

Exercise 3.1. Let $X$ be a variety of general type. Prove that a subvariety of $X$ passing through a general point is of general type.

Bogomolov proved in his paper "Families of curves on a surface of general type" that surfaces of general type satisfying $c_1^2 > c_2$ contain at most a finite number of rational and elliptic curves. It is unknown if the same holds true for an arbitrary surface of general type. Thus in the exercise above general point does not mean outside a closed subvariety. Instead it means outside a countable union of closed subvarieties. Of course, the author is considering all his varieties defined over $\mathbb C$.

Standard arguments reduces the exercise to the following

Exercise. Let $X$ be a variety of general type. If $Y \rightarrow T$ is a family of irreducible general type subvarieties of $X$ parametrized by a irreducible complex variety $T$ then the natural map $Y\to X$ is not dominant.

Theorem Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.

proof: By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.

Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.

proof: The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably many components. For each $P\in \mathbb{Q}[t]$, we have the universal family $$U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$$ Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. For any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form the Corollary above.