If complementary subspaces are almost orthogonal, is the same true for their orthogonal complements? Suppose that $W_1,W_2$ are complementary subspaces in a finite-dimensional vector space $V$ (so $W_1 + W_2 = V$ and $W_1 \cap W_2 = \{0\}$). Fix an inner product $\langle \cdot,\cdot\rangle$ on $V$ and further suppose that $W_1$ and $W_2$ are `almost orthogonal' with respect to this inner product i.e. if $w_1 \in W_1$ and $w_2 \in W_2$ are unit vectors, then $|\langle w_1, w_2 \rangle| \le \epsilon$.
It is true that $W_1^\perp$ and $W_2^\perp$ are complemented, but are they almost orthogonal? More precisely, can we estimate the quantity
$$
\sup_{v_i \in W_i^\perp, ||v_i|| = 1}  |\langle v_1, v_2 \rangle|
$$
and does it vanish as $\epsilon \to 0$.
 A: The head line question is answered with a plain yes.
And this yes remains true if $V$ is an infinite-dimensional Hilbert space.
It is assumed that $V=W_1\oplus W_2$, and the two complementary subspaces are necessarily closed (this merits special mention in the case $\dim V=\infty$).
Let $P_j$ denote the orthogonal projector (= idempotent and self-adjoint) onto $W_j$:
$$\sup_{w_j\in W_j\\ \|w_j\| = 1}\big|\langle w_1, w_2 \rangle\big|
 \;=\;\sup_{v_j\in V\\ \|v_j\| = 1}\big|\langle P_1v_1, P_2v_2 \rangle\big|
 \;=\;\sup_{v_j\in V\\ \|v_j\| = 1}\big|\langle v_1, P_1P_2v_2 \rangle\big|
 \:=\:\|P_1P_2\|\:=\:\epsilon\,<\,1$$
The last estimate is a non-obvious fact, cf
Norm estimate for a product of two orthogonal projectors . Only if
$W_1$ and $W_2$ are (completely) orthogonal one has $\epsilon=0\,$.  
Look at the corresponding quantity for the direct sum
$V=W_2{}^\perp\oplus W_1{}^\perp\,$: 
$$\sup_{w_j\in W_j{}^\perp\\ \|w_j\| = 1} \big|\langle w_2, w_1 \rangle\big|
 \;=\; \sup_{v_j\in V\\ \|v_j\| = 1} \big|\langle (\mathbb 1-P_2)v_2,
(\mathbb 1-P_1)v_1 \rangle\big| \:=\: \|(\mathbb 1-P_2)(\mathbb 1-P_1)\|$$
Because of $V=W_1\oplus W_2 = W_2{}^\perp\oplus W_2 = W_2{}^\perp\oplus W_1{}^\perp$
one can find unitaries $U_1:W_1\to W_2{}^\perp$ and $U_2:W_2\to W_1{}^\perp$, and thus define on $V$ the unitary operator
$$U: W_1\oplus W_2\xrightarrow{U_1\oplus\,U_2}W_2{}^\perp\oplus W_1{}^\perp$$
which respects the direct sums. Then $\mathbb 1-P_2=UP_1U^*$ and vice versa, hence
$$\|(\mathbb 1-P_2)(\mathbb 1-P_1)\|\;=\;\|UP_1U^*UP_2U^*\|\;=\;\|P_1P_2\|
= \epsilon\,.$$

Remark$\:\;\epsilon\,$ can be written as $\cos\gamma$, and $\gamma$ is interpreted as angle between the subspaces. This was the motivation for the post 
A "Crookedness criterion" for a pair of orthogonal projectors? .
A: Let $A_1 = {W_1}^\perp$ and $A_2 = {W_2}^\perp$.
Notice that $\dim A_1 = \dim W_2$ and that $\dim A_2 = \dim W_1$ so in order for them to be complementary we need only show that $A_1 \cap A_2 = \{0\}$.
Now, any $v\in V$ can be uniquely written as $v = w_1^v + w_2^v$, where $w_1^v \in W_1$ and $w_2^v\in W_2$.
If $a \in$ $A_1 \cap A_2$ then $a \perp v = w_1^v + w_2^v$ for all $v \in V$, and hence $a = 0$ as we sought to show.

Now, let $a \in A_1$ with $\lVert a\rVert = 1$ and write $a = w_1^a + w_2^a$.
Then
\begin{align}1
&=
\langle a, a \rangle
\\&=
\underbrace{\langle a, w_1^a \rangle}_0 + \langle a, w_2^a \rangle
\\&=
\langle w_1^a , w_2^a \rangle + \langle w_2^a , w_2^a \rangle\\{}
\end{align}
\begin{align}
&\implies
1 - \langle w_2^a , w_2^a \rangle = \langle w_1^a , w_2^a \rangle\tag{$*$}
\\&\implies
\big|1 - \langle w_2^a , w_2^a \rangle\big| = \big|\langle w_1^a , w_2^a \rangle\big| \leqslant \epsilon,
\end{align}
so that most of $\lVert a \rVert$ is concentrated on the $W_2$ component.
Now, of course, we can write
\begin{align}1
&=
\langle a, a \rangle
\\&=
\langle w_1^a+w_2^a, w_1^a+w_2^a \rangle
\\&=
\langle w_1^a, w_1^a\rangle
+ 2 \langle w_1^a, w_2^a\rangle
+ \langle w_2^a, w_2^a\rangle
\\&=
\langle w_1^a, w_1^a\rangle
+ 2 \Big(1 - \langle w_2^a , w_2^a \rangle\Big)
+ \langle w_2^a, w_2^a\rangle\tag{$**$}
\\&=
\langle w_1^a, w_1^a\rangle
+ 2
- \langle w_2^a, w_2^a\rangle
\end{align}
\begin{align}
&\implies
\langle w_1^a, w_1^a\rangle = \langle w_2^a, w_2^a\rangle - 1,
\end{align}
where we substituted $(*)$ into $(**)$.
This allows us to conclude that $\big|1 - \langle w_2^a , w_2^a \rangle\big| = \langle w_2^a, w_2^a\rangle - 1$, and hence
$$\left\{\begin{array}{}
0\leqslant \langle w_1^a, w_1^a\rangle \leqslant \epsilon\\
-\epsilon \leqslant \langle w_1^a , w_2^a \rangle \leqslant 0\\
1 \leqslant \langle w_2^a , w_2^a \rangle \leqslant 1+\epsilon
\end{array}\right.\tag{A}$$
Similarly, if $b\in A_2$ has $\lVert b \rVert = 1$, most of $\lVert b \rVert$ is concentrated on the $W_1$ component, with
$$\left\{\begin{array}{}
0\leqslant \langle w_2^b, w_2^b\rangle \leqslant \epsilon\\
-\epsilon \leqslant \langle w_1^b , w_2^b \rangle \leqslant 0\\
1 \leqslant \langle w_1^b , w_1^b \rangle \leqslant 1+\epsilon
\end{array}\right.\tag{B}$$
Finally, we can estimate $\langle a, b\rangle$ with
\begin{align}
\langle a, b\rangle
&=
\langle a, w_1^b + w_2^b\rangle
\\&=
\underbrace{\langle a, w_1^b\rangle}_{0} + \langle a, w_2^b\rangle
\\&=
\langle w_1^a, w_2^b\rangle + \langle w_2^a, w_2^b\rangle,
\end{align}
so that 
\begin{align}
|\langle a, b\rangle|
&\leqslant
|\langle w_1^a, w_2^b\rangle| + |\langle w_2^a, w_2^b\rangle|
\\&\leqslant
\epsilon + |\langle w_2^a, w_2^b\rangle| \tag{by almost orthogonality}
\\&\leqslant
\epsilon + \sqrt{\langle w_2^a, w_2^a\rangle\cdot\langle w_2^b, w_2^b\rangle}\quad\quad\quad\quad \tag{by Cauchy-Schwarz}
\\&\leqslant
\epsilon + \sqrt{(1+ \epsilon) \cdot \epsilon} \tag{by $(A)$ and $(B)$}
\end{align}
It follows that
$$
\sup_{v_i \,\in \,W_i^\perp\,\cap\, \partial B(0; 1)}  |\langle v_1, v_2 \rangle| \leqslant \epsilon + \sqrt{(1+ \epsilon) \cdot \epsilon}\,
,$$
so that it vanishes as $\epsilon \to 0$.
I'm not sure if the estimate is sharp, and moreover $V$ being finite dimensional did not come into play here.
We merely used the fact that $V = W_1 \oplus W_2$ and the 'almost orthogonality'.
