For any scheme and any closed subset, one can always consider the reduced induced subscheme structure. More generally, this works for any locally closed subscheme. For the first statement, one can consult Example 3.2.6 in Hartshorne. The second statement is in the book by Goertz-Wedhorn.
Consider the affine case $X = \text{Spec}(A)$. Then the closed subschemes of $X$ correspond to ideals $I \subset A$. If $\text{rad}(I) = \text{rad}(J)$, then $I$ and $J$ then $\text{Spec}(A/I)$ and $\text{Spec}(A/J)$ are (or more accurately, can be identified with) the same subsets of $\text{Spec}(A)$ (think about what it means to be a prime in $A/I$ and $A/J$), but if $I \neq J$ then they have different scheme structures. In this example, $\text{Spec}(A/\text{rad}(I))$ is the reduced induced closed subscheme structure on the set $\text{Spec}(A/I)$.
To make this really explicit, think about $X = \text{Spec}(\mathbb{C}[x])$ and consider the ideals $(x)$ and $(x)^2 = (x^2)$ in $\mathbb{C}[x]$. Both of them correspond to the point "$x = 0$" but the scheme structure (the "geometry") is different. You can see this by thinking about functions on $Z_1 = \text{Spec}(\mathbb{C}[x]/(x))$ vs. functions on $Z_2 = \text{Spec}(\mathbb{C}[x]/(x^2))$. For a function $f \in \mathbb{C}[x]$ (i.e. a function on $X$), when we look at it as a function on $Z_1$ we only see the constant term. But as a function on $Z_2$ we also get to see the linear term $a_0 +a_1x$.
For much more on this, I highly recommend the book "Geometry of Schemes" by Eisenbud and Harris.
Finally, you mention that you can't imagine a closed subvariety with more than one scheme structure. This is because a variety is (by definition) reduced, and thus there is only one scheme structure.
I figured I would make an answer out the comments to the first answer addressing a point which confused me when I first learned this stuff. A morphism $f:X\to Y$ (I have to write it in this direction or else I'll confuse myself) is said to be a closed immersion if $f$ induces a homeomorphism of $X$ onto a closed subset of $Y$, and $f^\sharp:\mathscr{O}_Y\to f_*(\mathscr{O}_X)$ is surjective.
In some references I've seen it is casually remarked that the second condition is equivalent to surjectivity of the map $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for all $x\in X$. But is this really trivial? No! This map, which might reasonably be called the stalk of the morphism $f$ at $x$, is not literally the same as the stalk of $f^\sharp$ at $f(x)$. Indeed, that is a map $f_{f(x)}^\sharp:\mathscr{O}_{Y,f(x)}\to(f_*\mathscr{O}_X)_{f(x)}$. In general, there is always a natural map $\varphi_x:(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$, but it isn't in general an isomorphism. The map $f_x^\sharp$ is equal to $\varphi_x\circ f_{f(x)}^\sharp$. So while it is standard that the map $f^\sharp$ of sheaves (on $Y$!) is surjective if and only if $f_y^\sharp:\mathscr{O}_{Y,y}\to (f_*\mathscr{O}_X)_y$ is surjective for all $y\in Y$, this does not obviously say anything about surjectivity of the maps $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for $x\in X$. If however $f$ is a homeomorphism onto a closed subset $f(X)\subseteq Y$, then the stalks of $f_*\mathscr{O}_X$ at points of $Y$ are easy to compute: they are zero at points outside of $f(X)$, and at a point $f(x)\in f(X)$, we have that the natural map $(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$ is an isomorphism. So in that case, surjectivity of each $f_x^\sharp$, $x\in X$, actually will imply surjectivity of $f_y^\sharp$, $y\in Y$, and hence of $f^\sharp$.
Without the condition that $f$ is a closed topological immersion on the underlying topological spaces, it is not going to be true that $f^\sharp$ is surjective if and only if $f_x^\sharp$ is surjective for all $x\in X$. To make this clearer, let's assume $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$, so $f=\mathrm{Spec}(\alpha)$ for $\alpha:A\to B$ a ring homomorphism. The stalk map of $f$ at $x=\mathfrak{q}\in\mathrm{Spec}(B)$ is the ring map $A_\mathfrak{p}\to B_\mathfrak{q}$, where $\mathfrak{q}=\alpha^{-1}(\mathfrak{p})$. In general, surjectivity of this map for all $\mathfrak{q}\in\mathrm{Spec}(B)$ does not imply surjectivity of $\alpha$ itself.
I think the simplest example that will illustrate this is when $B=A_g$ is a principal localization of $A$. Then in fact the stalk map in the previous paragraph is an isomorphism for every prime ideal of $A_g$ (the set of which are in natural bijection with the set of primes of $A$ not containing $g$, i.e. $D(g)$). But the localization map $A\to A_g$ (i.e. the map on global sections of $f$) is not usually surjective. Note that in this case $f$ is a homeomorphism onto the open subset $D(g)$ of $\mathrm{Spec}(A)$, but $D(g)$ is not generally closed in $A$.
I think maybe this illustrates why the first condition is important, and why, if one wants to think about surjectivity of $f^\sharp$ in terms of the stalks of $f$, $f_x^\sharp$, for $x\in X$, the topological condition is needed, and logically "precedes" the condition on $f^\sharp$.
Lastly, I should note that the maps which I have been calling the ``stalks of $f$," $f_x^\sharp$, for $x\in X$, are in fact the stalks of the map of sheaves on $X$ (in the usual sense) $f^\flat:f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$ corresponding to $f^\sharp$ under the adjunction between $f^{-1}$ and $f_*$. So surjectivity of all $f_x^\sharp$, $x\in X$, is logically equivalent to surjectivity of $f^\flat$. There is no reason to believe that $f^\flat$ is surjective if and only if $f^\sharp$ is, or even that there is an implication in either direction in general.
Best Answer
As a set, $V(\mathfrak{a})$ is equal to the image of the closed immersion $i:\operatorname{Spec}(A/\mathfrak{a}) \to \operatorname{Spec} A$. As a result, we can restrict $i$ to a homeomorphism $j:\operatorname{Spec}(A/\mathfrak{a})\to V(\mathfrak{a})$. We then make $V(\mathfrak{a})$ a scheme by transporting the scheme structure along $j$. Explicitly, the structure sheaf of $V(\mathfrak{a})$ is defined as $j_*\mathcal{O}_{\operatorname{Spec}(A/\mathfrak{a})}$.