Suppose that $A\in\mathbb{C}^{n\times n}$ is a hermitian matrix , then field of values is defined as $F(A)=\left\{x^{*}Ax:x^{*}x=1\right\}$. We can see that $F(A)$ is closed and bounded interval because it's a compact subset of $\mathbb{C}$ , also it contains all the eigenvalues of matrix $A$ (so it will also contain the min and max eigenvalue).
But how can we prove that $F(A)=[a,b]$ has $a=\lambda_{min}$ and $b=\lambda_{max}$??
The reason that $F(A)$ is an interval is not due to compactness but due to convexity.
Since $A$ is hermitian, $A=UDU^*$, with $U$ unitary and $D$ diagonal with the eigenvalues of $A$ in its diagonal. Then $F(A)=F(D)$. And $$ x^*Dx=\sum_jD_{jj}|x_j|^2=\sum_j\lambda_j|x_j|^2. $$ As $\sum_j|x_j|^2=1$, $$ \lambda_\min=\sum_j\lambda_\min|x_j|^2\leq\sum_j\lambda_j|x_j|^2\leq\sum_j\lambda_\max|x_j|^2=\lambda_\max. $$