Proposition. Let $k$ be an algebraically closed field. Then $V\subseteq k^n$ is finite if and only if $\Gamma(V)$ is a finite-dimensional $k$-vector space (where $\Gamma(V)=k[x_1,\dots,x_n]/I(V)$ is the affine $k$-algebra of polynomial functions on $V$)
For the "$\Rightarrow$" implication, with $V=\{u_1,\dots,u_r\}$, I know about the proof with the map $$\begin{align} \varphi:k[x_1,\dots,x_n] & \longrightarrow k^r\\ F & \longmapsto \varphi(F)=(F(u_1),\dots,F(u_r)) \end{align}$$ whose kernel is $I(V)$ and ... but the first "proof" I came up with is setting $$I(V)=I(\{u_1\}\cup\dots\cup\{u_r\})=I(u_1)\cap\dots\cap I(u_r)$$ with all $I(u_i)$ maximal by the Nullstellensatz, then $$\Gamma(V)=k[x_1,\dots,x_n]/I(V)=$$ $$=k[x_1,\dots,x_n]/(I(u_1)\cap\dots\cap I(u_r))\hookrightarrow$$ $$\hookrightarrow k[x_1,\dots,x_n]/I(u_1)\times\dots\times k[x_1,\dots,x_n]/I(u_r)=k^r$$ Here i use that $k[\dots]$ over a maximal ideal is a field and can be considered an extension of $k$ of at most countable degree (which makes it equal to $k$ as it is algebraically closed).
Is this proof correct? In essecence it looks the same but i find it more "geometrical" (dont know why).
Also, why can't we say that $\Gamma(V)$ is $r$-dimensional? What can happen so that it has lower dimension? If we translate/scale the whole set $V$ nothing should change as we can translate/scale the polynomials. Also, relative positions of the point shouldn't matter because in $I(V)$ there are polynomials of arbitrarily high degree that can adjust/interpolate (in some vague sense) the position of the points (the closer two of them are the higher the degree).
Thanks!