I'm reading Horn and Johnson's Matrix Analysis which presents a proof of Weyl's theorem that seems to be completely independent of Courant-Fischer. One of the exercises my teacher proposed is essentially to prove Weyl's theorem and he suggested using Courant-Fischer. Here's the exercise:
suppose $A,E \in \mathbb{C}^{n\times n}$ are hermitian with eigenvalues $\lambda_1 \geq \cdots\geq \lambda_n$, $\epsilon_1\geq\cdots\geq\epsilon_n$ respectively, and $B=A+E$ has eigenvalues $\mu_1\geq\cdots\geq\mu_n$. Prove that $\lambda_i + \epsilon_1 \geq \mu_i \geq \lambda_i + \epsilon_n$
My attempted proof goes as such:
$$\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*(A+E)v=\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \min_{v\in\mathcal{V}\\ \Vert v \Vert=1}(v^*Av+v^*Ev)$$ since the minimum of a sum is greater than the sum of the minimums, we have $$\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \min_{v\in\mathcal{V}\\ \Vert v \Vert=1}(v^*Av+v^*Ev)\geq\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \left(\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Av+\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Ev\right)$$ then we can claim that the minimum of $v^*Ev$ on $\mathcal{V}$ is greater than the minimum of $v^*Ev$ on $\mathbb{C}^{n\times 1}$, so we get $$\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \left(\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Av+\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Ev\right)\geq\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \left(\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Av+\min_{v\in\mathbb{C}^{n\times 1}\\ \Vert v \Vert=1}v^*Ev\right)$$ but the second minimum is $\epsilon_n$ by Rayleigh's theorem, so we're left with $$\max_{\mathcal{V}\leq\mathbb{C}^{n\times n}\\ \dim(\mathcal{V})=i} \left(\min_{v\in\mathcal{V}\\ \Vert v \Vert=1}v^*Av+\epsilon_n\right)=\lambda_i + \epsilon_n$$
which is the second inequality in the exercise. I suspect the other one can be obtained by an analogous argument. Sorry for the heavy notation, but is my argument correct? I feel it might have a hole somewhere.