I think this is a very interesting question. It gives another proof of the Segre embedding, makes it somehow more natural and puts it into a more general context, which I explain below.
I will just talk about the Segre embedding $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$. The general case of $\mathbb{P}(F) \times \mathbb{P}(G) \to \mathbb{P}(F \otimes G)$ is similar.
Let $x_0,...,x_n$ be the homogeneous coordinates of $\mathbb{P}^n$, $y_0,...,y_m$ the ones of $\mathbb{P}^m$ and $z_{00},...,z_{nm}$ the ones of $\mathbb{P}^{n+m+nm}$. The idea of the Segre embedding is $z_{ab} = x_a y_b$. Observe that with this equation we can interchange $a,c$ and $b,d$ in $z_{ab} z_{cd}$ without changing the result. Thus, consider the homogeneous ideal generated by the $z_{ab} z_{cd} - z_{cb} z_{ad}, z_{ab} z_{cd} - z_{ad} z_{cb}$ and let $I$ denote the corresponding ideal of the structure sheaf of $\mathbb{P}^{n+m+nm}$. Now I claim that $V(I)$ and $\mathbb{P}^n \times \mathbb{P}^m$ satisfy the same universal property. So these schemes are isomorphic, and we get a closed immersion $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$.
Let $X$ be a test scheme. Then $\hom(X,\mathbb{P}^n \times \mathbb{P}^m)$ is the set of tuples $(\mathcal{L}_1,\rho,\mathcal{L}_2,\tau)$, where $\mathcal{L}_1$ (resp. $\mathcal{L}_2$) is a line bundle on $X$ and $\rho_0,...,\rho_n$ (resp. $\tau_0,...,\tau_m$) are global generators of $\mathcal{L}_1$ (resp. $\mathcal{L}_2$).
On the other hand, $\hom(X,V(I))$ is the set of all tuples $(\mathcal{L},\sigma)$, where $\mathcal{L}$ is a line bundles on $X$ and $\sigma_{00},...,\sigma_{nm}$ are global generators of $\mathcal{L}$, which satisfy the relations $\sigma_{ab} \sigma_{cd} = \sigma_{cb} \sigma_{ad}$ and $\sigma_{ab} \sigma_{cd} = \sigma_{ad} \sigma_{cb}$.
With these descriptions, there is a very natural map $\hom(X,\mathbb{P}^n \times \mathbb{P}^m) \to \hom(X,V(I))$ given by $\mathcal{L} = \mathcal{L}_1 \otimes \mathcal{L}_2$ and $\sigma_{ab} = \rho_a \otimes \tau_b$. Actually this map makes sense in every abelian tensor category, but its bijectivity seems to be a special property of the abelian tensor category of quasi-coherent sheaves on a scheme (more generally, sheaves of modules on a ringed space). The usual proof uses the local explicit description of the classifying projective schemes, but you can do it directly by writing down an inverse map:
Let $\mathcal{L},\sigma$ be given as above. Construct a line bundle $\mathcal{L}_1$ as follows: For every $i$, put $X_{\rho_i} = \cup_j X_{\sigma_{ij}}$. On this open set, let $\mathcal{L}_1$ be generated freely by a symbol $\rho_i$. The cocycles $\rho_a / \rho_i$ are defined as $\sigma_{aj} / \sigma_{ij}$ for some $j$; the choice of this $j$ does not matter because of the relations for $\sigma$. Similarily we may define $\mathcal{L}_2$ with global generators $\tau_j$. It's easy to show that $\mathcal{L} \cong \mathcal{L}_1 \otimes \mathcal{L}_2$ with $\rho_i \otimes \tau_j$ corresponding to $\sigma_{ij}$.
Here is the answer for the Grassmannian of lines in $\mathbb{P}^3$. You can imitate this argumet in the general case of $k$-subspaces containing a fixed $(k-1)$-space or contained in a fixed $(k+1)$-space.
For any point $p\in\mathbb{P}^3$ be $\Sigma_{p}\subset\mathbb{G}(1,3)\subset\mathbb{P}^5$ be the locus parametrizing lines in $\mathbb{P}^3$ through $p$. Similarly, for any plane $H\subset\mathbb{P}^3$ be $\Sigma_{H}\subset\mathbb{G}(1,3)\subset\mathbb{P}^5$ be the locus parametrizing lines in $\mathbb{P}^3$ contained in $H$.
Let $u\in V$ be a vector representing $p\in\mathbb{P}^3$. Then the lines through $p$ are represented by $2$-vectors of the form $u\wedge v$. Let $\{u,u_1,u_2,u_3\}$ be a basis of $V$. Then we may write $v = \alpha u+\beta_1 u_1+\beta_2 u_2+\beta_3 u_3$, and
$$u\wedge v = \alpha_1(u\wedge u_1)+\alpha_2(u\wedge u_2)+\alpha_3(u\wedge u_3).$$
Therefore, lines through $p$ are represented by the points of the plane spanned by $u\wedge u_1$, $u\wedge u_2$ and $u\wedge u_3$.
Now, the lines contained in the plane $H\subset\mathbb{P}^3$, by duality corresponds to the lines in $\mathbb{P}^{3^{*}}$ through the point $H^{*}$. Therefore they are parametrized by a plane in $\mathbb{G}(1,3)$ by the part above.
Now, take a plane $\Pi$ in $\mathbb{G}(1,3)$ and three points $l,r,s$ in this plane that do not lie on the same line. Let $L,R,S\subset\mathbb{P}^3$ be the corresponding lines. Since the three lines joining $l$,$r$ and $s$ are on the same plane contained in $\mathbb{G}(1,3)$ they intersect and they are contained in $\mathbb{G}(1,3)$. Therefore, the lines $L,R,S$ intersect. We have two cases.
$L\cap R\cap S = \{p\}$. In this case $\Pi$ parametrizes lines in $\mathbb{P}^3$ through $p$.
$L$, $R$ and $S$ intersect in three distinct points. Let $u,v,w$ be three representative vectors for these three points. Then $L,R,S$ are represented by $v\wedge w$, $u\wedge w$ and $u\wedge v$. Then a point on the plane $\Pi$ is of the form
$$\alpha(v\wedge w)+\beta(u\wedge w)+\gamma(u\wedge v).$$
Therefore $L,R,S$ lie in the plane $\mathbb{P}(H)$, where $H = \left\langle u,v,w\right\rangle$.
Best Answer
Here is the argument I have written up in my thesis. It was suggested to me by my advisor Jarod Alper. We use the fact that a proper monomorphism is a closed immersion (EGA IV, 18.12.6). Furthermore, we will also use the functor of points perspective of the Grassmannian. Now because the Grassmannian is proper and projective space is separated, the Plucker embedding is proper (this is the property $\mathscr{P}$ exercise in Ravi Vakil's notes). Thus we just need to show it's a monomorphism. However, because :
it is enough to show the base change $$ \textbf{Gr}(d,n)_I \to \Bbb{P}\big(\bigwedge\nolimits^{\!d} \mathcal{O}_{\operatorname{Spec} \Bbb{Z}}^{\oplus n}\big)_I$$ is a monomorphism. One can view fact (2) above as the $\mathcal{O}_X$-module analog of the fact that a matrix is invertible iff its determinant is non-zero.
Now given a scheme $X$ and quotient $\mathcal{O}_{X}^{\oplus n} \twoheadrightarrow \mathcal{F}$, we may always replace it with a quotient $\mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ such that the composition $\mathcal{O}_X^I \hookrightarrow \mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ is the identity. This follows from the definition of when two quotients are equal in $\textbf{Gr}(d,n)_I(X)$. Henceforth, we will only work with quotients $M : \mathcal{O}_X^{\oplus n} \twoheadrightarrow \mathcal{O}_X^I$ of the following form: $M$ is a matrix (with coefficients in $\Gamma(X,\mathcal{O}_X)$) such that the $d$ columns of $M$ corresponding to the subset $I$ are equal to the columns of the $d \times d$ identity matrix. Now let $X$ be a scheme and suppose there are two quotients $M,M' \in \textbf{Gr}(d,n)_I(X)$ such that $\bigwedge^d M$ and $\bigwedge^d M'$ are equal. This by definition means we have a diagram
where the top, bottom rows are the identity and so is the left most column. Thus the displayed isomorphism is an honest equality. But now this means that the $d \times d$ minors of $M$ and $M'$ are equal and hence $M = M'$. In other words, the map $\textbf{Gr}(d,n)_I \to \Bbb{P}(\bigwedge^d \mathcal{O}_{\operatorname{Spec} \Bbb{Z}}^{\oplus n})_I$ is a monomorphism as claimed.