Let $k$ be an arbitrary field and $V\subseteq \mathbb{A}^n(k)$ be a Zariski-closed subset and $\Psi: V \to \text{Spec}(k[V]), x \mapsto m_x$, where $m_x \subseteq k[V]$ is the maximal ideal in the coordinate ring of the point $x \in V$. Is $\Psi$ a topological embedding?
If $k$ is algebraically closed, it is. But if $k$ is not algebraically closed, I am not sure how to show that it is continuous.
$\Psi$ is injective and it is $\text{Im} \Psi \cap V(I(V(I))) = \Psi(V(I))$ for any ideal $I \subseteq k[V]$. So, $\Psi$ is closed to its image. But why is $\Psi^{-1}(V(I))$ closed? If $k$ is algebraically closed, $I(V(I)) = \sqrt{I}$ by the Nullstellensatz and it follows. I think it should also work for non-algebraically closed fields that $\Psi$ is a topological embedding.
This is really general and has nothing to do with algebraic closure.
For any scheme $X$ whatsoever, the canonical unit $X\longrightarrow\operatorname{Spec}\mathcal{O}_X(X)$ is a genuine map of schemes and is in particular continuous. Take some $V(J)$ closed in the right hand side; the preimage is the set of all those $x\in X$ such that $J$ is contained in the vanishing ideal of $x$, such that if $f\in J$ then $f_x\in\mathfrak{m}_x$. Suppose $y$ is not in this set. Then, there is $f\in J$ such that $f_y$ is not in $\mathfrak{m}_y$; that is to say, $f$ is locally invertible at $y$. Unwinding the definition of stalk you see the existence of an open neighbourhood $V\ni y$ and a $g\in\mathcal{O}_X(V)$ such that $f|_V\cdot g=1$. For every $y'\in V$ it then follows $f_{y'}$ is invertible with inverse $g_{y'}$ hence $f_{y'}\notin\mathfrak{m}_{y'}$ hence $y'$ is not in our preimage.
That is to say, the preimage of some closed $V(J)$ is closed in $X$; a function is continuous iff. it pulls back closed sets to closed sets, so we are done.
Classically, (let's call our set $X$ rather than $V$) $k[X]=k[x_1,\cdots,x_n]/I$ for some radical ideal $I:=Z(X)$. Let $J$ be an ideal of this ring and consider $V(J):=\{\mathfrak{p}\subset k[x_1,\cdots, x_n]:J\subseteq\mathfrak{p}\}$; view $J$ as an ideal of $k[x_1,\cdots,x_n]$ that contains $I$. The map $X\to\operatorname{Spec}k[X]$ takes $x\to V(\{x\})$. The preimage of $V(J)$ is again just the set of all $x$ such that $f\in J$ implies $f(x)=0$. So, this preimage is just $Z(J)$, which is closed by definition. We don't need Nullstellensatz because I don't care whether or not the ideal $V(Z(J))$ is equal to $\sqrt{J}$ (sorry, the letter $V$ is used for so many things). I only care that the preimage of $V(J)$ under my map is $Z(J)$, which is automatic from the definition.