Ordered real eigenvalues are continuous on the real-rooted matrices

Idea. Roots of a polynomial depend continuously on its coefficients, in the sense that the number of roots in a small disc is stable. When all the roots are real, this stability pins down every ordered eigenvalue, because the discs are ordered along the real axis.

Step 1: setting up discs. Fix $A\in\mathcal S$ and let $\mu_1>\mu_2>\dots>\mu_r$ be its distinct eigenvalues, with multiplicities $m_1,\dots,m_r$ ($\sum m_j=n$). Let $\varepsilon>0$ be smaller than half of the smallest gap $\mu_j-\mu_{j+1}$. Then the closed discs $\overline D_j=\{|z-\mu_j|\le\varepsilon\}$ are pairwise disjoint, and
$$\lambda_k(A)=\mu_j\quad\text{for }m_1+\dots+m_{j-1}<k\le m_1+\dots+m_j.$$

Step 2: Rouché. The characteristic polynomial $\chi_A(z)=\det(zI-A)$ has no zero on the circles $C_j=\{|z-\mu_j|=\varepsilon\}$ (the only zeros are the $\mu_i$). So
$$m:=\min_j\min_{z\in C_j}|\chi_A(z)|>0.$$
The coefficients of $\chi_B$ are polynomials in the entries of $B$, hence depend continuously on $B$. The circles are compact and the degree is bounded by $n$, so $\sup_{z\in\bigcup C_j}|\chi_B(z)-\chi_A(z)|\to0$ as $B\to A$. Thus there is $\delta>0$ with
$$\|B-A\|<\delta\ \Longrightarrow\ |\chi_B(z)-\chi_A(z)|<m\le|\chi_A(z)|\quad\text{on every }C_j.$$
By Rouché's theorem, for such $B$ the polynomials $\chi_B$ and $\chi_A$ have the same number of zeros, counted with multiplicity, inside each open disc $D_j$: exactly $m_j$.

Step 3: using real-rootedness. Let moreover $B\in\mathcal S$. The $m_j$ zeros of $\chi_B$ in $D_j$ are real, so they lie in the interval $(\mu_j-\varepsilon,\mu_j+\varepsilon)$. Since $\sum_jm_j=n=\deg\chi_B$, there are no other zeros: all $n$ eigenvalues of $B$ lie in the union of the disjoint intervals $I_j=(\mu_j-\varepsilon,\mu_j+\varepsilon)$, and exactly $m_j$ of them in $I_j$.

Step 4: ordering. The intervals are ordered: every point of $I_1$ is larger than every point of $I_2$, and so on. Hence, in the decreasing list of eigenvalues of $B$, the first $m_1$ lie in $I_1$, the next $m_2$ lie in $I_2$, etc. Therefore, if $\lambda_k(A)=\mu_j$ then $\lambda_k(B)\in I_j$, i.e.
$$|\lambda_k(B)-\lambda_k(A)|<\varepsilon\qquad\text{whenever }B\in\mathcal S,\ \|B-A\|<\delta.$$
Since $\varepsilon$ can be taken arbitrarily small (with a corresponding $\delta$), $\lambda_k$ is continuous at $A$.

Check. Take $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ (so $\lambda_1=\lambda_2=0$) and $B_\varepsilon=\begin{pmatrix}0&1\\\varepsilon&0\end{pmatrix}$, $\varepsilon>0$. Its eigenvalues are $\pm\sqrt\varepsilon$, so $\lambda_1(B_\varepsilon)=\sqrt\varepsilon\to0$ and $\lambda_2(B_\varepsilon)=-\sqrt\varepsilon\to0$: continuous, as proved.

Remark. 1. The hypothesis $B\in\mathcal S$ is necessary: $\begin{pmatrix}0&-\varepsilon\\\varepsilon&0\end{pmatrix}$ has no real eigenvalue for $\varepsilon\ne0$ but converges to $0$, so "the largest real eigenvalue" is not a continuous function on $M_n(\mathbb{R})$.
2. The example above shows that $\lambda_k$ is continuous but in general not Lipschitz on $\mathcal S$: $|\lambda_1(B_\varepsilon)-\lambda_1(A)|=\sqrt\varepsilon=\sqrt{\|B_\varepsilon-A\|}$. On symmetric matrices, by contrast, Weyl's inequality gives $|\lambda_k(S)-\lambda_k(T)|\le\|S-T\|$.

Lesson. To prove continuity of an ordered quantity, first prove stability of the *count* in disjoint regions (Rouché), then let the ordering of the regions do the rest.