Let $\mathbb{A}^1$ be the affine line (considered as algebraic variety) endowed with action by $\mathbb{G}_m$ via $t \cdot a = ta$.
If we form a set theoretical quotient $\mathbb{A}^1/\mathbb{G}_m$ then it consists of exactly two points/orbit classes, namely $\{0\}$ and $\mathbb{A}^1 \backslash \{0\}$.
My QUESTION is what is the concretee reason why this (naive) quotient cannot have structure of an algebraic variety? (remark: this observation motivates to define the quotient in categorial way in order to acheve a quotient in right category).
Attempts: the "point" $\mathbb{A}^1 \backslash \{0\}$ is dense in $\mathbb{A}^1/\mathbb{G}_m$ so every element of the coordinate ring $K[\mathbb{A}^1/\mathbb{G}_m]$ vanishing in $\mathbb{A}^1 \backslash \{0\}$ would vanish everywhere. Does it lead to a contradiction?
Futhermore: Is there a pure topological argument wrt structure of Zariski topology which provides the desired result?
The set theoretical quotient doesn't make sense, in the sense that the action map $\mathbb{G}_m\times_k \mathbb{A}^1_k\to \mathbb{A}^1_k$ is not a well-defined map on the underlying topological space.
Let's assume that you mean that you want $\mathbb{A}^1_k/\mathbb{G}_m$ to be two points $X=\{\eta,s\}$ with topology $\{X,\{\eta\}\}$. Then, this topological space certainly is the underlying space of a scheme. For example, if $A$ is any DVR then $\mathrm{Spec}(A)$ is homeomorphic to $X$.
But, let's show that there is no natural $k$-scheme structure on $X$ and a surjective map $\mathbb{A}^1_k\to X$ which is $\mathbb{G}_m(k)$-invariant where we assume that $k=\overline{k}$ for simplicity. Note $X$ is necessarily affine since $s$ has a neighborhood which is affine, but its only neighborhood is $X$. Thus, $X=\mathrm{Spec}(R)$ for some $R$. Note that if we have a map $\mathbb{A}^1_k\to X$ which is surjective (or just dominant, they're the same thing here) then we know that the map $R\to k[x]$ has kernel $I$ contained in the nilradical of $R$. Note though that by assumption we have that the image of $k[x]$ lands in the $\mathbb{G}_m(k)$-invariants of $k[x]$ which is just $k$. Since $R$ is a $k$-algebra this forces the image of $R$ to be $k$. Thus, $R/I\cong k$. But, since $I$ is contained in the nilradical of $R$ we know that $|\mathrm{Spec}(R/I)|=|\mathrm{Spec}(R)|$ (where $|S|$ is the underlying space of $S$) but $R/I=k$ so this is a contradiction.