Kevin Lin's answer regarding the meaning of closed points is quite reasonable, especialy in the case when the scheme in question underlies a classical variety. I want to add some additional remarks and examples for thinking about more general schemes.
Here are some tautological remarks: recall that a point $x$ in a scheme $X$ is called a specialization of $y$ if $x$ lies in the Zariski closure of $y$ (and $y$ is called a generalization of $x$). So tautologically, a closed points is one that cannot be specialized any further (just as a generic point cannot be generalized any further). What does specialization really mean:
ring theoretically, it means taking the image under a homomorphism; so if $\mathfrak p$ and $\mathfrak q$ are prime
ideals of a ring $A$, then $\mathfrak q$ is a specialization of $\mathfrak p$ in $\text{Spec} A$ if and only if $\mathfrak q$ contains $\mathfrak p$, i.e. if $A/\mathfrak p$ surjects onto $A/\mathfrak q$. It is perhaps best to think of an example: say $A$ is $\mathbb C[x,y]$,
$\mathfrak p$ is the prime ideal gen'd by $(x-1)$ and $\mathfrak q$ is the prime (actually maximal ideal) gen'd by $(x-1,y)$. Then in $A/\mathfrak p$, we have "specialized" the value of $x$ to equal $1$ (because we have declared $x-1 = 0$) but $y$ is still a free variable. When we pass to the further quotient $A/\mathfrak q$, we have specialized both $x$ and $y$: $x$ is specialized to $1$ and $y$ is specialized to $0$. At this point, we can't specialize any more; technically, this is because $\mathfrak q$ is a maximal ideal of $A$,
so a closed point of $\text{Spec} A$; intuitively, it is because both $x$ and $y$ have now both been "specialized" to actual numbers, and so we can't specialize any further.
But suppose now we set $B = \mathbb Z[x,y]$, and take $\mathfrak p$ and $\mathfrak q$ to be the same, i.e. gen'd by $x-1$ and by
$(x-1,y)$ respectively. Then $\mathfrak q$ is not maximal; there is more capacity for specialization.
How is this? Well, $x$ and $y$ are now taking values in $\mathbb Z$ (rather than the field $\mathbb C$) and so we can also reduce both $x$ and $y$ modulo some prime, say $5$; this gives a prime ideal $\mathfrak r = (x-1,y,5)$ in $B$ containing $\mathfrak q$. Now $\mathfrak r$ is maximal, and so we are done specializing.
So if you have a scheme that is finite type over $\mathbb Z$, the closed points will correspond to
"actual points", in Kevin's terminology, but defined over finite fields. The points of the scheme whose coordinates are integers, say, will not be closed. One has the choice of thinking them of them as "actual points" which nevertheless can be specialized further by reducing modulo primes, or as subvarieties rather than "actual points", by identifying them
with their Zariski closures (for a picture of this, see the drawing of Mumford that Kevin links to).