We'll show that the $A\mapsto ( \lambda_1, \ldots, \lambda_n)$ from hermitian $n\times n$ matrices to the eigenvalues in decreasing order is Lipschitz. ( One can choose specific norms on both sides but that won't affect the statement).
First, let us show that if $A \preceq B$ ( the semidefinite order, meaning $B- A \succeq 0$ ) then the $k$th eigenvalue of $A$ is $\le $ the $k$-th eigenvalue of $B$, that is $\lambda_k(A) \le \lambda_k(B)$, for all $1 \le k \le n$. Indeed, there exists an $n-k+1$-dimensional subspace $V_1$ of $\mathbb{C}^n$ such that for every vector $x$ in that subspace we have $\langle A x, x \rangle \ge \lambda_k(A) \cdot ||x||^2$, and there exists a $k$-dimensional subspace $V_2$ of $\mathbb{C}^n$ such that for every vector $x$ in that subspace we have $\langle B x, x \rangle \le \lambda_k(B) \cdot ||x||^2$. These two subspaces have a non-zero intersection, let $0 \ne x \in V_1 \cap V_2$. We have
$$\lambda_k(A) \cdot ||x||^2 \le \langle A x, x \rangle \le \langle B x, x \rangle \le \lambda_k(B) \cdot ||x||^2$$ and so $\lambda_k(A) \le \lambda_k(B)$.
Now, recall an important ( not hard to prove ) result : if $C$ is an $n\times n$ hermitian matrix so that $c_{ii} > (\ge) \sum_{j, j\ne i} | c_{ij}|$ for all $1 \le i \le n $ ( diagonally dominant) then $C \succ (\succeq) 0$.
Consider the norm $C \mapsto |C| = \max ( |c_{ij}|)$ on hermitian matrices. From the above we conclude that if $|C| \le \frac{\epsilon}{n}$ then
$\epsilon \cdot I \pm C \succeq 0$. As a consequence, if $|A-B| \le \frac{\epsilon}{n}$ then $\epsilon \cdot I \pm (A-B) \succeq 0$. From $\epsilon \cdot I + A \succeq B$ we get, using the preceding, $\epsilon + \lambda_k(A) \ge \lambda_k(B)$. Similarly $\epsilon + \lambda_k(B) \ge \lambda_k(A)$. Therefore
$$|\lambda_k(A) - \lambda_k(B)| \le n \cdot |A-B|$$
for all $1 \le k \le n$.
${\bf Added:}$ The constant $n$ is the best possible. Indeed, consider the $n\times n$ matrix $A$ with all entries $1$. $A$ has $n$ as eigenvalue ( eigenvector all entries $1$) and $|A| = |A-0| = 1$.
As for the non-dfferentiability : the matrix$t \left(\matrix{t&0\\0& -t}\right)$ has largest eigenvalue $|t|$.
Best Answer
I don't understand most of your question, but the reason that $\lambda_\max(A(x))\le t$ is equivalent to $A(x)-tI\le 0$ is clear, because $\lambda$ is an eigenvalue of $A(x)$ if and only if $\lambda-t$ is an eigenvalue of $A(x)-tI$. (More specifically, this is because $Ax=\lambda x$ iff $(A(x)-tI)v=(\lambda -t)v$, where $v$ denotes an eigenvector.) Therefore, $\lambda_\max(A(x))\le t$ if and only if all eigenvalues of $A(x)-tI$ are nonpositive, i.e. if and only if $A(x)-tI\le0$.