It is a simple and repeated application of $\min$ and $\max$ operators.

$$v^*(A+B)v=v^*Av+v^*Bv\le v^*Av+\|B\|_{op},\,\forall v\in R^n\wedge |v|=1.$$
Given $V$ where $\dim(V)=i$,
$$\min_{u\in V,|u|=1}u^*(A+B)u\le v^*(A+B)v\le v^*Av+\|B\|_{op},\,\forall v\in V\wedge |v|=1,$$
then
$$\min_{u\in V,|u|=1}u^*(A+B)u\le \min_{v\in V,|v|=1}v^*Av+\|B\|_{op}\le \max_{\dim(V)=i}\min_{v\in V,|v|=1}v^*Av+\|B\|_{op}$$
and
$$\max_{\dim(U)=i}\min_{u\in V,|u|=1}u^*(A+B)u\le \max_{\dim(V)=i}\min_{v\in V,|v|=1}v^*Av+\|B\|_{op}.$$
In other words
$$\lambda_i(A+B)\le\lambda_i(A)+\|B\|_{op}.$$
Similarly, we can prove
$\lambda_i(A)-\|B\|_{op}\le\lambda_i(A+B)$
and reach the desired inequality.