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I'm looking for an easy argument to see that the set of maximal flags of a vector space is a projective variety.

This seems to be somewhat intuitive, since 1-dimensional subspaces correspond to points in projective space and we also have that the more general Grassmannians (= the set of flags of minimal length) is projective. However, the projective embedding (the Plücker embedding) is already quite complicated.

Even without giving an explicit embedding, is there an easy way to see that the set of maximal flags is projective?

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    If you believe that Grassmannians are projective, embed the flag variety into a product of Grassmannians by sending the flag to each of its component subspaces. Then show that the image of this embedding is closed. – 2017-01-28
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    @QiaochuYuan That's a nice idea. Do you mean that for a flag $0 = V_0 \subset V_1 \subset \dotsb \subset V_{n-1} \subset V_n = V$ you send it to the point $(V_1, \dotsc, V_{n-1})$ in $\prod_{0 < i < n} \operatorname{Gr}_i (V)$? Might it be easier to send it to $(V_1, \dotsc, V_{n-1}) \in \prod_{0 – 2017-01-28

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I think @QiaochuYuan's comment can be turned into an answer.

The set $S$ of maximal flags in an $n$-dimensional vector space $V$ is isomorphic to $\mathbb P^1 \times \dotsb \times \mathbb P^{n-1}$.

Each point in $S$ is a maximal flag $0 = V_0 \subset V_1 \subset \dotsb \subset V_{n-1} \subset V_n = V$ with $\dim V_i = i$ so that $\dim V_i / V_{i-1} = 1$. Thus each point $(v_1, \dotsc, v_{n-1}) \in \mathbb P^1 \times \dotsb \times \mathbb P^{n-1}$ defines a maximal flag by choosing $V_i$ to be the complement of the one-dimensional subspace corresponding to $v_i \in \mathbb P^i \simeq \mathbb P (V_{i+1})$.

Now use that the product of projective spaces/varieties is projective.