For each $N \in \mathbb{N}$, Let $S_N$ be the set of positive definite matrices on $\mathbb{R}^{N\times N}$, and let W be a nonempty, convex and compact subset of $\mathbb{R}^N$. For $Q$ in $S_N$, let
$x_Q = argmin_{x \in W}\ \ \ x^TQx$
Note that $x_Q$ exists and is unique under the stated assumptions. Let $A=\{x_Q: Q \in S_N\}$. Is $A$ a convex subset of $\mathbb{R}^N$?
EDIT: Hitch Alisson has found a counterexample using a convex, compact subset of $\mathbb{R}^N$ with positive volume in $\mathbb{R}^N$. What if the elements of $W$ must satisfy some linear restriction? For example, let $a \in \mathbb{R}$ be given and fix the first element of every vector in $W$ to equal $a$.
END OF EDIT
If so, consider the following generalisation. Let $X$ be a $(K \times N)$ matrix with $K\geq N$ that is full rank. Define
$y_Q = argmin_{y \in W}\ \ \ y^TX^TQXy$
and let $B^X=\{y_Q: Q \in S_K\}$. Is $B^X$ a convex set in $\mathbb{R}^N$, under the restriction on $W$ mentioned in the edit?
Do you mean a convex set of ${\mathbb{R}}^{n}$? Then not necessarily.
Take W to be a ball outside the unit sphere then only the points on the surface of the ball (facing the origin) can be minimizers of a positive definite quadratic form.
If $x$ is in the interior of $W$ then $\exists \, y \in W$ co-linear with $x$ of smaller norm (i.e. such as $y = \alpha.x$ and $\alpha < 1$) in which case $y^{T}Qy < x^{T}Qx$. The surface of the ball is not a convex set of ${\mathbb{R}}^{n}$.