Inclusion of closed subschemes. 
Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$? 

This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, \mathcal{O}_Y)$ and $(Z, \mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, \mathcal{O}_X)$. Then $\mathcal{O}_Y$ and $\mathcal{O}_Z$ are quotients of the structure sheaf $\mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is, 
$$\mathcal{O}_Y = \mathcal{O}_X/\mathcal{I}$$
and 
$$\mathcal{O}_Z = \mathcal{O}_X/\mathcal{J} \, .$$
Then, to say that $Y \subset Z$, we would like to say that $|Y| \subset |X|$ as topological spaces, and that $\mathcal{O}_Y$ is a sub sheaf of $\mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $\mathcal{J}$ is a subsheaf of $\mathcal{I}$. Would this suffice?
 A: For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).
Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Y\subset X$ and quasi-coherent ideal sheaves $\mathcal J\subset \mathcal O_X$.
The corresponding items are written  $Y=V(\mathcal J)$, resp. $\mathcal J=\mathcal I_Y$.
And then the answer to your question is  given by the biblical simple equivalence:

$$Y\subset Z\iff   \mathcal I_Y \supset \mathcal I_Z$$ 

Edit: a simple but edifying example.
Let $(A,\mathfrak m)$ be a discrete valuation ring, $X=\operatorname {Spec}A=\{M,\eta\}$ its associated affine scheme and $U=\{\eta\}$ its non trivial open subset.      
a) The only non zero ideals of $A$ are the powers $\mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $\mathcal I_n$ define the closed subschemes $Y_n=V(\mathcal I_n)\subset X$.
These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1=\{M\}$ and they constitute the collection of all the closed subschemes  $\subsetneq X$.
Notice that $Y_n$ is non reduced for $n\gt 1$ and that $Y_n\subsetneq Y_m$ for $n \lt m$.
b) However $\mathcal O_X$ has another sheaf of ideals $\mathcal J$ defined by $\mathcal J(X)=A$ and  $\mathcal J(U)=\{0\}$.
This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.
