This follows from the following characterisation of the eigenvalues of symmetric matrices.

\[\lambda_k(A) = \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2}\]

Here the minimum is taken over all subspaces $U$ of $\mathbb{R}^n$ of dimension $n-k+1$. (I chose to write $\lambda_K(A)$ rather than $\lambda_k^A$.) The proof of this uses the fact that symmetric matrices have an orthonormal basis of eigenvectors.

Now,

\[\begin{aligned}

\lambda_k(A) &= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2} \\

&= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^T(A-B+B)x}{\|x\|_2} \\

&= \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \left[ \frac{x^TBx}{\|x\|_2} + \frac{x^T(A-B)x}{\|x\|_2}\right] \\

&\leqslant \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \left[ \frac{x^TBx}{\|x\|_2} + \|A-B\|\right] \\

&= \lambda_k(B) + \|A-B\|

\end{aligned}\]

So $\lambda_k(A) - \lambda_K(B) \leqslant \|A-B\|$. Analogously we have $\lambda_k(B) - \lambda_K(A) \leqslant \|A-B\|$

This inequality is a particular case of

Weyl's Inequalities