The spectral radius is continuous

Idea. $\rho(A)$ is the largest modulus of a root of $\chi_A$. Moving the coefficients a little moves each root a little, and so moves the largest modulus a little.

Step 1: set-up. Let $A_k\to A$. The coefficients of $\chi_{A_k}$ converge to those of $\chi_A$ (they are polynomial in the entries), so $\eta_k:=\max_j|c_j(A_k)-c_j(A)|\to0$; discard finitely many terms so that $\eta_k\le1$. Let $C$ be the constant of Problem 81 for $p=\chi_A$.

Step 2: upper bound. Let $\lambda$ be an eigenvalue of $A_k$ with $|\lambda|=\rho(A_k)$. By Problem 81, there is an eigenvalue $\mu$ of $A$ with $|\lambda-\mu|\le C\eta_k^{1/n}$, so
$$\rho(A_k)=|\lambda|\le|\mu|+C\eta_k^{1/n}\le\rho(A)+C\eta_k^{1/n}.$$

Step 3: lower bound. Let $\mu$ be an eigenvalue of $A$ with $|\mu|=\rho(A)$. By the converse part there is an eigenvalue $\lambda$ of $A_k$ with $|\lambda-\mu|\le C\eta_k^{1/n}$, so
$$\rho(A_k)\ge|\lambda|\ge|\mu|-C\eta_k^{1/n}=\rho(A)-C\eta_k^{1/n}.$$

Step 4: conclusion. $|\rho(A_k)-\rho(A)|\le C\eta_k^{1/n}\to0$.

Not Lipschitz. $A_\varepsilon=\begin{pmatrix}0&1\\\varepsilon&0\end{pmatrix}$ has characteristic polynomial $t^2-\varepsilon$, so $\rho(A_\varepsilon)=\sqrt\varepsilon$ while $\rho(A_0)=0$ and $\|A_\varepsilon-A_0\|=\varepsilon$. The ratio $\sqrt\varepsilon/\varepsilon\to\infty$. (Numerically: $\varepsilon=10^{-4}$ gives $\rho=0.01$.) For the $n\times n$ Jordan block with a corner entry $\varepsilon$, $\rho=\varepsilon^{1/n}$.

Remark. On the other hand $\rho$ *is* Lipschitz on the set of symmetric (more generally normal) matrices, where $\rho(A)=\|A\|$ (Weyl, Problem 88).