If a subspace $V_1$ is close to a subspace $V_2$ of the same dimension, is $V_2$ close to $V_1$? Given two vector subspaces $V_1,V_2$  of an infinite-dimensional normed space $X$,  we say that $V_1$ is $\epsilon$-close to $V_2$, 
if for every $v_1 \in V_1$, $\|v_1\|=1$, there exist $v_2 \in V_2$ such that $\|v_1-v_2\|<\epsilon$.
Now, suppose that $(V_n)_{n \ge 0}$ is a sequence of $k$-dimensional subspaces ($k < \infty$) of $X$, such that $V_n$ is $\epsilon_n$-close to $V_0$, where $\epsilon_n \to 0$. 

Is it true that $V_0$ becomes arbitrarily close to $V_n$ when $n \to \infty$?

In other words, I am asking whether the closeness relation is "asymptotically-symmetric".
The "equal dimensions" assumption is necessary: Otherwise $V_n$ could be strict subspaces of $V_0$.
Edit:
gerw proved that the answer is positive whenever $X$ is a Hilbert space.
The question remains whether or not the answer stays positive if the norm is not induced by an inner product.
 A: After giving a flawed counterexample, I give a proof that works (at least) in Hilbert spaces.
Let us assume that $V_n$ is $\varepsilon$-close to $V$ with $\varepsilon < 1$. Let us denote by $S$ the unit ball in our Hilbert space $X$. Moreover, let $P$ be the orthogonal projection onto $V$.
By $\varepsilon$-closedness, we have $\| v - P v\| \le \varepsilon$, hence, $\| P v \| \ge 1-\varepsilon > 0$ for all $v \in V_n \cap S$, i.e., the map $Q : V_n \cap S \to V \cap S$ defined via $Qv := Pv/\|Pv\|$ is continuous and odd. Since $V_n \cap S$ and $V \cap S$ are essentially $n-1$-spheres, $Q$ should be surjective, this should follow, e.g., from Borsuk-Ulam theorem. Now, its inverse $Q^{-1}$ gives you a nice map from $V \cap S$ to $V_n$. This shows that $V$ is $\delta$-close to $V_0$ for some $\delta$. Moreover, we get $\delta \to 0$ as $\varepsilon \to 0$.
Here is some explanation for the should. I was not able to come up with a reference, so I might miss something obvious:
Let $f : S^n \to S^n$ be a continuous, odd map, i.e., $f(-x) = -f(x)$. Here, $S^n$ is the $n$-sphere. We claim that $f$ is surjective. Otherwise, there is $x \in S^n$ which is not in the image of $S^n$. Since $f$ is odd, $-x$ does not belong to the image, too. Now, let $g : S^n \to \mathbb R^n$ be the orthogonal projection onto the orthogonal complement of $x$ within $\mathbb R^{n+1}$. Then, $g\circ f : S^n \to \mathbb R^n$ is a continuous, odd map and $0$ does not belong to the image of $g \circ f$. This is a contradiction to Borsuk-Ulam.
