Notation:
$[n] = \{1,\dotsc,n\}$.
For symmetric $n \times n$ matrices $A,B$, $A \succ B$ and $A \succeq B$ means that $A-B$ is positive definite / positive semidefinite.
For a symmetric matrix $A$, $\lambda_{\min}(A)$ and $\lambda_{\max}(A)$ denote its minimal and maximal eigenvalues.
Claim:
If $n \ge 2$, and $u_1, \dotsc, u_n, w, a \in \mathbb{R}^d$ are unit vectors such that $u_j^T u_k < 0$ for every $j \neq k$, and $u_j^T w > 0$ for every $j$, then $\sum_j |u_j^T a| (u_j^T w) < 1$.
Proof:
If $u_j^T a = 0$ for every $j$, or $u_j^T w = 0$ for every $j$, then this is trivial.
Otherwise we can replace $a$ by a scalar multiple of its orthogonal projection onto $\sum_j \mathbb{R} u_j$, and the same for $w$.
So we assume that $a, w \in \sum_j \mathbb{R} u_j$, and we can then also assume that $\mathbb{R}^d = \sum_j \mathbb{R} u_j$.
We prove by induction on $n$.
Let $G = (u_j^T u_k)_{j,k \in [n]} \succeq 0$ be the Gram matrix of $(u_j)_{j=1}^n$, and let $M = I-G$.
Then $M_{j,j} = 0$ for every $j$, and $M_{j,k} > 0$ for $j \neq k$.
Using the Perron-Frobenius theorem (https://en.wikipedia.org/wiki/Perron-Frobenius_theorem) for $\epsilon I + M$ with $\epsilon > 0$, we get that its maximal eigenvalue $\epsilon + \lambda_{\max}(M)$ has multiplicity $1$, and it has a corresponding eigenvector in $\mathbb{R}_{>0}^n$, and $\epsilon + \lambda_{\min}(M) \ge -(\epsilon + \lambda_{\max}(M))$.
Taking $\epsilon \to 0$, we get $\lambda_{\min}(M) \ge -\lambda_{\max}(M)$.
So $\lambda_{\max}(G) + \lambda_{\min}(G) \le 2$, and we also see that the eigenvalue $\lambda_{\min}(G)$ of $G$ has multiplicity $1$.
We could also use the same argument for $(u_1,\dotsc,u_n, -w)$ instead of $(u_1,\dotsc,u_n)$, and get an $(n+1) \times (n+1)$ Gram matrix $G' \succeq 0$, where $\lambda_{\min}(G') = 0$ has multiplicity $1$, so $\operatorname{rank}(u_1,\dotsc, u_n) = \operatorname{rank}(G') = n$.
So $u_1, \dotsc, u_n$ are linearly independent (so $d = n$), and thus $G \succ 0$ and $0 < \lambda_{\min}(G)$ and $\lambda_{\max}(G) < 2$, so $-I \prec M \prec I$.
We cannot have $\lambda_{\min}(G) = \lambda_{\max}(G) = 1$, because then $M = 0$.
So $\lambda_{\min}(G) \in (0,1)$.
Note that $G^{-1} = (I-M)^{-1} = I+M+M^2+\dotsm$ (this converges, because $-I \prec M \prec I$), so $(G^{-1})_{j,k} > 0$ for every $j,k$.
So by the Perron-Frobenius theorem, $G^{-1}$ has a unique unit length eigenvector in $\mathbb{R}_{>0}^n$, and the corresponding eigenvalue is $\lambda_{\max}(G^{-1})$ (this will be used later).
We fix linearly independent unit vectors $(u_j)_{j \in [n]}$, and choose unit vectors $a, w$ in $\mathbb{R}^d$ so that $S = \sum_j |u_j^T a| (u_j^T w)$ is maximal, only requiring $u_j^T w \ge 0$ for every $j$.
If $u_j^T w = 0$ or $u_j^T a = 0$ for some $j$, then we are done by induction.
So let $u_j^T w > 0$ and $u_j^T a \neq 0$ for every $j$.
Then $S > 0$.
Let $a = \sum_j \alpha_j u_j$ and $w = \sum_j \beta_j u_j$.
Then $1 = \|a\|^2 = \alpha^T G \alpha$, $1 = \|w\|^2 = \beta^T G \beta$, and $u_j^T a = (G \alpha)_j \neq 0$ and $u_j^T w = (G \beta)_j > 0$.
Let $D$ be the diagonal matrix such that $D_{j,j} = \operatorname{sgn}(u_j^T a) = \operatorname{sgn}((G \alpha)_j)$ for every $j$.
Then $S = \sum_j (G \alpha)_j D_{j,j} (G \beta)_j = \alpha^T G D G \beta$.
Because $(\alpha, \beta)$ give a local maximum for $S$, the differential of the map $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^3$, $(\alpha, \beta) \mapsto (\alpha^T G \alpha, \beta^T G \beta, \alpha^T G D G \beta)$ does not have full rank at our $(\alpha, \beta)$.
So there is an $(A, B, C) \in \mathbb{R}^3 \setminus \{0\}$ such that
$$
2 A G \alpha + C G D G \beta = 2 B G \beta + C G D G \alpha = 0.
$$
Then $0 = \alpha^T (2 A G \alpha + C G D G \beta) = 2 A + C S$ and $0 = \beta^T (2 B G \beta + C G D G \alpha) = 2 B + C S$, so $A = B = -\frac{C}{2} S$, where $C \neq 0$.
So $GDG \beta = S G \alpha$ and $GDG \alpha = S G \beta$.
So if $x_0 = G \alpha$ and $y_0 = G \beta$, then $(x_0,y_0)$ satisfies $1 = x_0^T G^{-1} x_0 = y_0^T G^{-1} y_0$, $GD x_0 = S y_0$, $GD y_0 = S x_0$.
The idea is to study the pairs $(x,y)$ that satisfy these equations, because these could maybe also give the same maximal $S$, so in some cases we could vary $(\alpha, \beta)$ even if $S$ is already maximal.
Note that $(x_0, y_0)$, $(-x_0, -y_0)$, $(y_0, x_0)$, $(-y_0, -x_0)$ all satisfy these equations
Idea: If $(x,y)$ is on the ellipse centered at $0$ going through these four points, then it should also satisfy these equations.
In detail:
Let $p = x_0+y_0$, $q = x_0-y_0$, $P = p^T G^{-1} p$, $Q = q^T G^{-1} q$.
Then $GD p = S p$, $GD q = -Sq$ (so $p,q$ are eigenvectors of $GD$), $p^T G^{-1} q = 0$, $P, Q \ge 0$ and $P + Q = 4$.
Let us take $x = sp + tq$ and $y = sp - tq$ for $s, t \in \mathbb{R}$.
Then $GD x = S y$ and $GD y = S x$ hold, and $x^T G^{-1} x^T = y^T G^{-1} y^T = P s^2 + Q t^2$, so we need this to be $1$.
Then $GDy=Sx$, so $x^T D y = S x^T G^{-1} x = S$.
So if $P s^2 + Q t^2 = 1$, then $(x,y) = (sp + tq, sp - tq)$ gives the same maximal $S$.
If $p = 0$ or $q = 0$ (so $a = \pm w$):
Then $D = (\operatorname{sgn}(u_j^T a))_{j=1}^n = \pm I$ and $G y_0 = S y_0$, so $G^{-1} y_0 = S^{-1} y_0$.
Here $(y_0)_j = u_j^T w > 0$ for every $j$, so $y_0$ is a Perron-Frobenius eigenvector for $G^{-1}$, thus $S^{-1} = \lambda_{\max}(G^{-1})$, so $S = \lambda_{\min}(G) < 1$.
Now let $p,q \neq 0$.
Then $P, Q > 0$, so we get an ellipse.
For $s = t = \frac{1}{2}$ we get $(x,y) = (x_0, y_0)$, for $s = t = -\frac{1}{2}$ we get $(x,y) = (-x_0, -y_0)$.
So moving along the ellipse, we can get from $(a,w)$ to $(-a, -w)$, without changing $S$.
So there will be a first point where some $u_j^T a$ or $u_j^T w$ becomes $0$, and then we are done by induction.
