This is probably something standard and I just don't know where to look (so a reference would be just as appreciated as an answer), but...
Let $\mathfrak{g}$ be a finite dimensional semisimple Lie algebra over a field $k$. People sometimes define the flag variety $\mathcal{B}$ to be the set of Borel subalgebras of $\mathfrak{g}$, and then endow $\mathcal{B}$ with the structure of a variety by regarding it as a subset of the grassmannian of $\dim(\mathfrak{b})$-dimensional subspaces of $\mathfrak{g}$, where $\mathfrak{b}$ is some fixed Borel subalgebra (for instance, see page 129 in section 3.1 of Chriss and Ginzburg's book Representation Theory and Complex Geometry).
Why is $\mathcal{B}$ a closed subset of the grassmannian?
In this my post I gave a contruction of the flag manifolds as the quotient spaces of $G=GL(n,\mathbb{C})$ over a parabolic subgroup $P$; if $P$ is a minimal parabolic subgroup (that is a minimal element in the set of parabolic subgroups of $G$), that is if $P$ is a Borel subgroup of $G$, $G_{\displaystyle/B}$ is the full flag manifold $\mathcal{F}(n;1,\dots,n)\equiv\mathcal{F}$; wich is a closed subset of $G(1,n)\times G(2,n)\times...\times G(n,n)$, where $G(k,n)$ is the Grassmannian of the $k$-planes of $\mathbb{C}^n$.
One checks:
As reference I suggest: Nicolas Perrin - Linear algebraic groups, on line lecture notes, chapters 7 and 8.
Addendum. In other way, let $B$ be a Borel subgroup of $G$ and let $P_1,\dots,P_k$ be the maximal parabolic subgroup of $G$; one can prove that the canonical map \begin{equation*} G_{\displaystyle/B}\to G_{\displaystyle/P_1}\times\dots\times G_{\displaystyle/P_k} \end{equation*} is an embedding.