If $V$ is a $k$-variety of dimension $1$, any proper closed irreducible subset of $V$ is a single point. 
If $V$ is a $k$-variety of dimension $1$, how I can prove that any proper closed irreducible subset of $V$ is a single point?

Intuitively, I see the reason, but I could not show it. I used this as an argument in a proof, so I need to know how to prove.
My definition for dimension is the largest integer $n$ such that there is a chain
$$Y_{0} \subset Y_{1} \subset ... \subset Y_{n} \subseteq V$$
of irreducible $k$-subvarieties of $V$.
 A: Well, let's see. Let $W\subset V$ be a proper closed (nonempty) irreducible subset of $V$. So we have a point $x\in W$.
The trick is to observe that the irreducible subsets of $V$ correspond bijectively to closures of points. Note that the closure of $x$, call it $Z$, is a closed subset of $W$. It's even irreducible since the closure of an irreducible subset is irreducible and a point is irreducible.
Now $Z=W$ since, if this were not true we would have proper containments $$Z\subset W\subset V$$ which contradicts that $V$ has dimension 1. It just remains to show that $x=Z$.
If $Z\neq x$, then there's another point $y$ in $Z$. But, by the same argument as the above, we find that the closure of $y$ must be $Z$. So now we have two points, $x\neq y$ both having closure $Z$.
A point $x$ whose closure is the subset $Z$ is usually called a generic point of $Z$, and it's not too hard to show that these generic points are unique. This will be the last step in the proof: once we show that generic points are unique, we'll have shown $x=Z=W$.
Now we need to use some algebra (because for arbitrary topological spaces, there is no reason for a generic point to be unique. I.e. the two point set with the topology given by the whole space and the empty set gives an example). First, every point of our variety $V$ is contained in some affine open set. This is true also for the point $x$, so let $U=\mathrm{Spec}(R)$ be an open affine set which contains $x$.
Since $x$ is a point of $U$, it is given by some prime ideal $\mathfrak{p}_x\subset R$; similarly $y$ is given by some prime ideal $\mathfrak{p}_y$. We need to look at the closure of these points, which I'll recall is the intersection $$V(\mathfrak{p}_x)=\bigcap V(J)=V(\mathfrak{p}_y)$$ where $J$ runs over all ideals containing $\mathfrak{p}_x$ (or $\mathfrak{p}_y$ since we're assuming this is the same). Now from this equality we want to extract some information about the ideals and their containments relative to one another. To do this we use the operation $I(-)$ which takes closed subsets to ideals.
We have $$r(\mathfrak{p}_x)=I(V(\mathfrak{p}_x))=I(V(\mathfrak{p}_y))=r(\mathfrak{p}_y)$$ where $r(-)$ means "the radical ideal of --". But, radicals of ideals are intersections of primes containing them and of course $\mathfrak{p}_x\supset \mathfrak{p}_x$ (sim. with $y$) so that the above equality actually reads $$\mathfrak{p}_x=r(\mathfrak{p}_x)=r(\mathfrak{p}_y)=\mathfrak{p}_y.$$
And this finishes the proof.
