I wanted to show that in general, T has a unique fixed point and to determine that fixed point of T such that $Tf = f$
$T : (C[0,1], \left\lVert.\right\rVert_{inf} \rightarrow C[0,1], \left\lVert.\right\rVert_{inf} )$, with $\left\lVert f\right\rVert_{inf} = \sup_ {x\in [0,1]}{|f(x)|}$ be defined by
$(Tf)(x) = \int_0^1\frac{1}{2}(xy-1)(f(y) + 1)\mathrm{d}y$
Thanks for the help.
Existence and Uniqueness:
\begin{align} & \lVert (Tf)(x) - (Tg)(x) \rVert \\ = & \bigg\lVert \int_0^1\frac{1}{2}(xy-1)(f(y) + 1)dy - \int_0^1\frac{1}{2}(xy-1)(g(y) + 1)dy \bigg\rVert \\ = & \sup_{x \in [0,1]} \bigg| \int_0^1\frac 12 (xy-1)(f(y)-g(y))dy \bigg| \\ ≤& \sup_{x \in [0,1]} \int_0^1 \bigg| \frac 12 (xy-1)(f(y)-g(y))\bigg| dy \\ =& \sup_{x \in [0,1]} \frac 12 \int_0^1 |xy-1| \bigl|f(y)-g(y)\bigl|dy \\ ≤ & \frac 12 \int_0^1 \bigl|f(y)-g(y)\bigl|dy && x,y\in[0,1] \implies |xy-1|≤1\\ ≤ & \frac 12 \int_0^1 \sup_{y \in [0,1]}|f(y)-g(y)|dy \\ = & \frac 12 \sup_{y \in [0,1]}|f(y)-g(y)| \\ = & \frac 12 \lVert f-g \rVert \end{align}
Hence, this map is a contraction with constant $\frac 12$, and so by the Contraction Mapping Theorem, a unique fixed point exists.
Construction:
\begin{align} (Tf)(x) = f(x) \implies f(x) & = \int_0^1 \frac 12 (xy-1)(f(y)+1)dy \\ & = \frac x2 \int_0^1y(f(y)+1)dy - \frac 12 \int_0^1 (f(y)+1)dy \\ & = Ax+B \end{align}
for some constants $A$ and $B$.
i.e. $f$ must be a linear function of the form $f(x)=Ax+B$. Plug this back into the above:
\begin{align} & (Tf)(x) \equiv f(x) \\ \implies & Ax+B \equiv \int_0^1 \frac 12 (xy-1)(Ay+B+1)dy \\ \implies & Ax+B \equiv \biggl( \frac A6 + \frac B4 + \frac 14 \biggl)x + \biggl(-\frac A4 - \frac B2 - \frac 12 \biggl) \end{align}
Comparing coefficients, we get
$$\begin{pmatrix} \frac 56 & -\frac 14 \\ \frac 14 & \frac 32 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} \frac 14 \\ -\frac 12 \end{pmatrix}$$
This gives $A = \dfrac{4}{21}$ and $B = -\dfrac{23}{63}$, and so the unique fixed point is
$$f(x) = \dfrac{4}{21} x -\dfrac{23}{63}$$