Statement of the problem:
Let $i: X \to Y$ be the inclusion of a closed subvariety. Then $i$ is a finite morphism.
Here's my attempt. Please verify it or correct it. Suggestions for improvement are welcome:
I should first show that $i$ is an affine morphism, i.e. there exists an open covering $Y = \bigcup_i U_i$ by affine sets $U_i$ such that the open subsets $i^{-1}(U_i)$ are affine.
Since $Y$ is a variety, by definition it can be covered by a finite number of open affine sets $U_i$, i.e. $Y=\bigcup_{i=1}^n U_i$. $X$ is a closed subvariety of $Y$ and $i^{-1}(U_i)=X \cap U_i$ is a closed subvariety of $U_i$ which is affine; therefore, $X \cap U_i$ being a closed subspace of an affine variety must be affine itself. This shows that $i^{-1}(U_i)$ is affine for any $i$ and hence $i$ is an affine morphism.
Now I want to show that each $k[X\cap U_i]$ is finitely generated as a $k[U_i]$-module (i.e. it is of finite type). I do not understand this part well but here's my attempt:
By definition, for an open set $U$ in $X$, a regular function on $U$ is a function of the following form: there is an open cover $U=\cup (X\cap V_{\alpha})$ where the $V_{\alpha}$'s are open subsets of $Y$ such that $f(y)=g_{\alpha}(y)$ on $X \cap V_{\alpha}$ where $g_{\alpha}$ is regular on $V_{\alpha}$.(This is the definition given in Kempf's Algebraic Varieties)
Assuming that definition is indeed well-defined, it should be independent of how we cover $U$. But $X \cap U_i$ is an open subset of $X$ and surely it can be covered by itself. Therefore, a regular map on $X \cap U_i$ comes from a regular map on $U_i$. This shows that if $f \in k[X \cap U_i]$, then $f = g_i \times 1$ where $g_i \in k[U_i]$ and therefore, $k[X \cap U_i]$ is generated by the constant function $1$ which is regular on $X \cap U_i$. So, as a $K[U_i]$-module, $K[X \cap U_i]$ is generated by scalar multiples of one element (namely the constant function $1$) and hence is finitely generated.
This is all a bit laboured. We can reduce to the affine case: $X=\textrm{Spec}\,R$ and $Y=\textrm{Spec}\,S$. For $i:X\to Y$ to be a closed immersion (the inclusion of a closed variety is a special case), then the corresponding ring map $S\to R$ is a surjection: basically $R=S/I$ for the ideal $I$ for which $X=V(I)$. Therefore $R$ is a finitely-generated $S$-module, and so $i$ is a finite morphism.