Spectrum of a Subspace of Matrices

This is not true. Consider $$ A(t) = \begin{pmatrix} 0 & t & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} . $$ Then $\det (A(t)+1) = 1$, so $-1\notin S$, but clearly the spectrum of $A(t)$ is not constant (for example, $0$ is in the spectrum if and only if $t=0$).

To expand on Christian Remling's answer a bit: in his example, setting $det(A(t) - \lambda) = 0$ becomes $(1-\lambda)t+\lambda^3=0$, and solving for $t$ gives $t = x^3/(x-1)$ -- so for any $\lambda \in \mathbb{C}$, we have $x \in S$ by choosing this $t$, with the exception of $\lambda=1$.

In general for an affine subspace of dimension $d$, we can define the space by $A(t_1,t_2,\dots t_d)$, and all entries in $A$ are linear in the $t_i$. Then $\det(A(t) - \lambda)$ is an $n$th degree polynomial in all the $t_i$ and $\lambda$. It is possible that the determinant has no dependence on any $t_i$, in which case we have a finite spectrum. Otherwise, we have a dependence on the $t_i$. Then $\det(A(t)-\lambda)=0$ has a solution in $t_1$ whenever at least one of the non-constant coefficients is nonzero. (In the above example, as long as the linear coefficient $1-\lambda\neq 0$.) Since the leading coefficient (that is not a constant 0 polynomial) is an $n-1$th degree polynomial in $\lambda$, it is zero at a most $n-1$ different places, and this is at most $n-1$ places that the spectrum can fail to be continuous. The spectrum only fails to be continuous when all of the coefficients in the $t_1$ polynomial are zero, so this is an upper bound. So, theorem:

For an affine subspace of $n\times n$ matrices, the spectrum is either finite (of size at most $n$), or it is $\mathbb{C}\setminus K$, where $K$ is a finite set of exceptions (of size at most $n-1$).