Finite Morphism of Schemes is Proper Let $f: X \to Y$ a finite morphism of schemes (therefore $f$ affine and the quasi koherent $\mathcal{O}_Y$-algebra $\mathcal{A}= f_*(\mathcal{O}_X)$ such that $X = Spec(\mathcal{A})$ is a finite $\mathcal{O}_Y$-algebra)
Therefore practically that means that for every affine open set $V = Spec(S) \subset Y$ the $S=\Gamma(V,\mathcal{O}_Y)$-algebra $A= \Gamma(V,\mathcal{A})$ is finite
I want to show that then $f$ is proper.
Here my attempts:
I want to use following valuative criterion for properness:
$f$ is proper (so  separated, of finite type, and universally closed) if and only if in every diagram
$$
\require{AMScd}
\begin{CD}
Spec(F)  @>{g}  >> X   \\
@VViV  @VVfV   \\
Spec(R) @>{t}>> Y 
\end{CD}
$$
for every discrete valuation ring $R$ and $F = Frac(R)$ there exist $l:Spec(R) \to X$ such that $a = l \circ i$ and $t = f \circ l$ holds.
The spectrum of DVR $R$ has the structure $Spec(R) = \{\sigma, \eta\} = \{\sigma\} \cup Spec(F)$ with generic point $ \eta$ and unique maximal ideal $\sigma$.
Since $Spec(F)$ is a sigleton, we can reduce the problem to affine case $X= Spec(A), Y = Spec(S)$ with $A$ finite $S$-algebra.
For underlying topological spaces it's obviously how to define $l: Spec(R) \to X = Spec(A)$:
Let $s_0:= t(\eta), s_1 := t(\sigma)$. Obviously $s_0 \subset s_1$ as prime ideals. Because $= s_0= t \circ i (\eta) = f \circ g(\eta)$, we conclude that $a_0:= g(\eta)$ is lying over $s_0$. Since $f$ is finite morphism the going up theorem holds, therefore there exist a $a_1$ lying over $s_1$. 
Define $l$ set theoretically by setting $l(\eta)= a_0, l(\sigma) := a_1$.
Now to my problem: How to see that $l$ defined in this way provides a morphism of schemes, not only a map which commutates with the commutative square?
 A: As you have remarked, it is sufficient to look at the case where $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} S$ are both affine. In general, when defining morphisms between affine schemes, it is easier to define the corresponding map between the rings. So instead of trying to define construct directly a scheme map from $\operatorname{Spec} R$ to $X$, it is easier to construct a ring map from $A$ to $R$. This is the same thing, since we have a canonical isomorphism $\operatorname{Hom}_{Schemes}(\operatorname{Spec} R, X) = \operatorname{Hom}_{Rings}(A, R)$. 
Turning the valuative criterion that you quoted into commutative algebra, we see that what you want to show is equivalent to the following proposition. 
Proposition Let $A$ be a finite $S$-algebra. Then for any discrete valuation ring $R$ with fraction field $F$ and any a commutative diagram 
$$
\require{AMScd}
\begin{CD}
S  @>{t}  >> R   \\
@VVfV  @VViV   \\
A @>{g}>> F 
\end{CD}
$$
there is a ring map $A \to R$ that fits in the diagram. 
Of course, we usually think of $R$ as a subring of $F$, so saying that there is a morphism $A \to R$ as in the diagram really just means that the map $g$ has image contained in $R \subset F$. 
To prove the above proposition, we just note that the image $g(A)$ of $A$ will be finite and therefore integral over $R$. So $g(A)$ is contained in the integral closure of $R$ in $F$. But $R$ is a discrete valuation ring, and such rings are integrally closed in their fraction field. This gives us $g(A) \subset R$ as needed.  
