Let $H$ be a Hilbert space and $\langle , \rangle$ its inner product. Let $X : $ be a self-adjoint linear operator and let $F_X : H \to \mathbb{R}$ be defined as $F_X(a) = \langle a , Xa \rangle$. Let $U \subseteq \mathbb{R}$ be an open set. Is $F_X^{-1}(U)$ a Borel set (in the Borel algebra of the topology induced by the norm of the Hilbert space)?
From what I understand, if $X$ is bounded then it is continuous, therefore $F_X$ is continuous, therefore $F_X^{-1}(U)$ is an open set and therefore a Borel set.
What can I say in the unbounded case?
May we assume that $X$ is a linear operator? If so then one has boundedness if and only if continuity as you describe. The missing step then would be show that that your self-adjoint operator is bounded using the Uniform Boundedness Principle. The rest follows by using the characterisation of a continuous map as one where the pre-image of open sets are open.
Let me elaborate on the UBP part: Our goal is to show that $X$ is bounded as this implies the boundedness of $F_X$ and we're done. That is to say, we wish to show that $\sup_{\vert\ x \vert \leq 1}\vert\vert Xx \vert\vert < \infty$. To do this we start by defining the following family of linear operators $T_x:H\rightarrow H$ defined $T_x(y) := <Xx,y>_H$. One may show that for any choice of $x$ that these are bounded, in particular we have: $\sup_{\vert y \vert \leq 1}\vert\vert T_x y \vert\vert = \sup_{\vert y \vert \leq 1}<Xx,y> = \vert\vert Xx\vert\vert$. Similarly, one may show that $\sup_{\vert x \vert \leq 1}\vert\vert T_x y \vert\vert = \sup_{\vert x \vert \leq 1}<Xx,y> = \sup_{\vert x \vert \leq 1}<x,Xy> = \vert\vert Xy\vert\vert$. Having satisfied the conditions of the UBP we conclude that $\vert X \vert := \sup_{\vert x \vert \leq 1}\vert Xx \vert = \sup_{\vert x\vert\leq 1, \vert y \vert \leq 1}\vert T_xy\vert < \infty$