Space of finite dimensional subspaces is separable In Bernard Maurey's paper "A Note on Gowers' Dichotomy Theorem" at the top of the 7th page, the following fact is stated that I'm not able to prove:
Let $X$ an infinite dimensional separable normed Space. Then the space of finite dimensional subspaces of $X$ is separable with respect to the Hausdorff metric of spheres.
For a normed space $V$, denote by $S(V)$ the unit sphere in $V$ and let $d(x,S(V)) = inf\{\|x-y\|: y \in S(V)\}$. Then we define the Hausdorff distance of spheres between to subspaces $U$ and $V$ of a normed space $X$ by $\delta(U,V) = max\{sup\{d(x,S(U)): x \in S(V)\}, sup\{d(y,S(V)): y \in S(U)\}\}$.
The natural way of approaching this problem would be: Choose an arbitrary finite dimensional subspace $M$ of $X$ and choose a normalized basis. Then approximate each basis element by appropriate elements of a previously fixed dense subset of $S(X)$. On a Hilbert space this will give you the desired result by using an orthonormal basis and the Cauchy-Schwarz-inequality. But on a general normed space you will have to argue using the equivalence of norms on finite dimensional spaces. This approach seems to be much more subtle because the constants of equivalence of $M$ and it's approximating space differ.
 A: It is a general fact that the space $\mathcal{K}(X)$ of non-empty compact subsets of a separable metric space $(X,d)$ is separable with respect to the Hausdorff metric. In fact, if $D$ is a countable dense subset of $X$ then the set $\mathcal{F}(D)$ of non-empty finite subsets of $D$ is countable and dense in $\mathcal{K}(X)$.
To see this, let $K$ be any compact subset of $X$. Choose finitely many $\varepsilon$-balls with centers in $K$ whose union covers $K$ and let $x_1,\dots,x_n \in D$ be points of these balls. Then the $\varepsilon$-neighborhood of $K$ in $X$ contains $\{x_1,\dots,x_n\} \in \mathcal{F}(D)$ by construction. Conversely, the $2\varepsilon$-balls around the points $x_1,\dots,x_n$ contain $K$, so that $\delta(K,\{x_1,\dots,x_n\}) \leq 2\varepsilon$.
Now observe that sending a finite-dimensional space to its unit sphere yields an isometric embedding of the space of finite-dimensional subspaces of $X$  into $\mathcal{K}(S(X))$ where $S(X)$ is the unit sphere of the Banach space $X$. Recall that a subspace of a separable metric space is itself separable.
This approach also yields an explicit countable dense subset of the space of finite-dimensional subspaces of $X$: Choose a countable dense subset $D$ of the unit sphere of $X$ and consider the set of finite-dimensional spaces generated by finite subsets of $D$.
