Interesting question! I think it is possible. My construction is not very linear algebraic, though!
The existence of your $w$ tells us that all of $S$ (and therefore all of the convex hull of $S$) lies strictly in one "half" of $\Bbb R^d$. Therefore, the convex hull of $S$ has a face that's closest to the origin (possibly a face with a large number of dimensions!). Then we can just use the orientation of this face to get a new vector $\hat w$, which defines a plane parallel to this face. By convexity, the convex hull of $S$ is still on one side of this plane.
More formally: The convex hull of $S$ is a compact set, because $S$ is finite. Therefore there is a point $x$ in the convex hull closest to the origin. We may write $x = \sum_i \lambda_i x_i$ as a convex combination of the elements $x_i$ of $S$ (so $0 \le \lambda_i \le 1$ and $\sum_i \lambda_i = 1$). Note that $x \ne 0$, because $w \cdot x > 0$ (by using this expression for $x$). I'll show that $x$ is the $\hat w$ that you seek.
Indeed, suppose there was some $y \in S$ with $x \cdot y \le 0$. This will be absurd because if we perturb a little bit in the direction of $y$, we'll get a new point in the convex hull closer to the origin. Indeed, consider the vector $x + \varepsilon (y - x)$ for $0 < \varepsilon \le 1$. Clearly this is still in the convex hull. But also,
\lVert x + \varepsilon (y - x) \rVert^2
&= \lVert (1 - \varepsilon)x + \varepsilon y \rVert^2 \\
&= (1 - \varepsilon)^2 \lVert x \rVert^2 + \varepsilon^2 \lVert y \rVert^2
+ 2 \varepsilon(1 - \varepsilon) x \cdot y \\
&= \lVert x \rVert^2 + \varepsilon((\varepsilon - 2)\lVert x \rVert^2 + \varepsilon\lVert y \rVert^2
+ 2 (1 - \varepsilon) x \cdot y)
Taking $\varepsilon$ small enough, we see that the term $(\varepsilon - 2)\lVert x \rVert^2 + \varepsilon\lVert y \rVert^2 + 2 (1 - \varepsilon) x \cdot y$ will be negative (since it tends to $-2\lVert x \rVert^2 + 2x \cdot y < 0$ as $\varepsilon$ tends to $0$. This is where we need $x \ne 0$). So we've found something closer to $0$, which is a contradiction! So we had $x \cdot y > 0$ and we're done.
In the last part, equivalently we worked out that the derivative of the function $\varepsilon \mapsto \lVert x + \varepsilon (y - x) \rVert^2$ is negative at $\varepsilon = 0$.
The only part where we used the existence of $w$ was to guarantee $x \ne 0$, and indeed this answer shows that if $0$ is not in the convex hull of $S$, then there is such a vector $\hat w$ in the convex hull. Therefore in fact the existence of any such $w$ is equivalent to $0$ not being in the convex hull of $S$.