For the interested: To prove the existence of such an operator, let $x \in \mathcal{H}$ and define the following function $f_x(y):= \int_0^1 (F(t),x)(F(t),y)\,dt$. We will show that this function is bounded, namely:
\begin{align*}
|f_x(y)| &= |\int_0^1 (F(t),x)(F(t),y)\,dt| \\
&\leq \int_0^1 |(F(t),x)(F(t),y)|\,dt \\
& = \int_0^1 |(F(t),x)||(F(t),y)|\,dt \\
&\leq |x|\int_0^1 |F(t)|^2\,dt |y| \quad \text{(Cauchy-Schwarz)} \\
&\leq|x| \max_{t \in [0,1]}|F(t)|^2 |y|
\end{align*}
which gives that $f$ is a bounded operator since $f$ is continuous in the compact interval $[0,1]$, so attains a maximum. Due to the fact that $f$ is bounded, by the Riesz-Fr\'echet representation theorem there exists a unique $z \in \mathcal{H}$ such that $f(x) = (z,y)$. We then define $T$ such that $T(x) = z$. This shows that there exists a unique $T$ such that for all $x,y \in \mathcal{H}$ we have $(Tx,y) = \int_0^1 (F(t),y)(F(t),x)\,dt$.
Now we want to show that $T \in B(\mathcal{H})$, so that it is linear and bounded. For linear, let $x_1,x_2 \in \mathcal{H}$ and $\lambda,\mu \in \mathcal{R}$. Then:
\begin{align*}
(T(\lambda x_1 + \mu x_2),y) &= \int_0^1 (F(t),\lambda x_1 + \mu x_2)(F(t),y)\,dt \\
&= \int_0^1 (\lambda(F(t),x_1) + \mu(F(t),x_2))(F(t),y)\,dt \\
&= \lambda\int_0^1 (F(t),x_1)(F(t),y)\,dt + \mu\int_0^1 (F(t),x_2)(F(t),y)\,dt \\
&= \lambda(Tx_1,y) + \mu(Tx_2,y) = (\lambda Tx_1,y) + (\mu Tx_2,y)
\end{align*}
which proves linearity. Now for the boundedness of $T$, take $x \in \mathcal{H}$ then we have by Cauchy-Schwarz that:
\begin{align*}
|Tx|^2 &= (Tx,Tx) \\
&= \int_0^1 (F(t),x)(F(t),Tx)\,dt \\
&= \int_0^1 (F(t),x)\int_0^1 (F(t),x)(F(t),F(t))\,dt\,dt \\
&= \int_0^1 (F(t),x)^2 |F(t)|^2 \,dt \\
&\leq \left(\int_0^1 |F(t)|^4\,dt\right) |x|^2 \leq \max_{t \in [0,1]}|F(t)|^4 |x|^2
\end{align*}
which in turn implies that $|Tx| \leq \max_{t\in[0,1]}|F(t)|^2 |x|$. Therefore $T \in B(\mathcal{H})$.
We will now show that $T$ is a positive operator. Let $x \in \mathcal{H}$, then:
\begin{align*}
(Tx,x) &= \int_0^1 (F(t),x)(F(t),x)\,dt \\
&= \int_0^1 (F(t),x)^2\,dt \geq 0
\end{align*}
since $(F(t),x)^2 \geq 0$ for all $x \in \mathcal{H}$ and all $t \in [0,1]$. For the self-adjoint condition, this is easily seen since we are working on a real Hilbert space. Then:
\begin{align*}
(Tx,y) &= \int_0^1 (F(t),x)(F(t),y)\,dt \\
&= \int_0^1 (F(t),y)(F(t),x)\,dt \\
&= (Ty,x) = (x,Ty).
\end{align*}
As the last required property, we will show that $T$ is a compact operator. Look at the following:
\begin{align*}
\sum_{n=1}^\infty |Te_n|^2 &= \sum_{n=1}^\infty (Te_n,Te_n) \\
&= \sum_{n=1}^\infty \int_0^1 (F(t),e_n)\int_0^1 (F(t),e_n)(F(t),F(t))\,dt\,dt \\
&= \int_0^1 |F(t)|^2 \sum_{n=1}^\infty (F(t),e_n)(e_n,F(t)) \,dt \\
&= \int_0^1 |F(t)|^2 |F(t)|^2\, dt \quad \text{(by Parseval relation Q3.24)} \\
&\leq \max_{t \in [0,1]}|F(t)|^4
\end{align*}
which is bounded by the compactness of $[0,1]$, which means that $T$ is a Hilbert-Schmidt operator and therefore compact.