I am looking for alternate proofs for the convexity of the field of values.
In Topics in matrix analysis by Horn and Johnson they define the field of values of an $n\times n$ matrix $A$ as $$F(A) = \left\{x^*Ax\mid x^*x = 1\right\}.$$ It turns out that $F(A)$ is always convex and in the book they show this by first showing that $F(A)$ is convex for any $A$ if $$F\begin{pmatrix}0 & a\\b& 0\end{pmatrix}$$ is convex for any real and nonnegative $a,b$. They then show that such a set always is a (possibly degenerate) closed ellipse.
While I understand the provided proof I still don't feel like I have an intuitive understanding for why $F(A)$ would be convex. In the book they mention that "there are other proofs in the literature that are based upon more advanced concepts from other branches of mathematics" but they don't provide any specific references.
Any alternate argument for why $F(A)$ would be convex is appreciated.
Have you tried to work through problem 7 of section 1.3 (convexity of the field of values) of Horn's book ? It proposes a variant of Hausdorff's proof of the convexity. First it reduces (by shift and rotation) the property to be proved to a simple case:
(P) If $0$ and $a\in \mathbb R, a>0$ are in $F(A)$, the segment $[0,a]$ is containted in $F(A)$.
Now, for $x^*x=y^*y=1$ with $x^*Ax=0$ and $y^*Ay=a$, the proof constructs a continuous curve $z(t)$ in the unit sphere $\{ w\ |\ w^*w=1\}$ such that $z(0)=0$ and $z(1)=y$ and $z(t)^*A z(t)$ is real for all $t$. Now, the intermediate value theorem proves the claim (P). The construction uses the the decomposition of $A$ in a hermitian and an anti-hermitian part and diagonalisation of the latter. Maybe you find this proof more accessible, as it directly verifies the convexity.
Just a hint to why the proofs are not so easy after all: $F(A)$ is defined as the image of a (non-convex) unit-sphere in $\mathbb C^n$ under a quadratic form. Thus, a $S^{2n-1}$ is mapped to a convex set in $\mathbb C$. The construction is specific for quadratic forms, the structure theory of such forms enters (as above: decomposition, diagonalisation).
Only in the real (hermitian A) case the statement is obvious: The connected sphere is mapped to an interval in $\mathbb R$, which is convex.
p.s.
You may have a look at Chandler Davis' (https://cms.math.ca/openaccess/cmb/v14/cmb1971v14.0245-0246.pdf) proof. He makes the geometric part of the argument more transparent, representing $F(A)$ as the linear image of a 2-sphere in $\mathbb R^2$, which is necessarily convex. The essential observation is: $$ x^*Ax= Tr (x^* Ax)= Tr(A xx^*)\; .$$In the two-dimensional case the set of all $xx^*$ is a two-sphere, the linear map in question is $Tr (A \cdot)$.