This might be a very obvious, but I will ask it nevertheless:
We know from the minimax principle that the $i$-th largest value of a symmetric matrix $A \in M_{n,n}(\mathbb{R})$ can be written as
$$ \lambda_i = \max_{U \subset \mathbb{R}^n,\\ \dim(U)=i} \min_{u \in U \\ ||u||=1} u^t A u.$$
I know that the minimum is well-defined since $u \mapsto u^t A u$ is continuous and $\{u \in U : ||u||=1 \} $ is compact, as it is obviously bounded and closed as the intersection of two closed subsets $U$ and $S^1 = \{ x \in \mathbb{R}^n : ||x||=1 \}$.
But I do not understand why exactly the maximum exists. Could you please explain that to me? Thank you!
We can see the existence of the maximum from the proof of Courant-Fischer itself.
By the Spectral Theorem, we have unit eigenvectors $e_1, \ldots, e_n$ which corresponds to the eigenvalues $\lambda_1, \ldots, \lambda_n$, and $\lambda_1\geq \cdots \geq \lambda_n$.
Let $U$ be a subspace or $\mathbb{R}^n$ whose dimension is $i$. Consider the space $W$ spanned by $\{e_i, e_{i+1}, \ldots, e_n\}$. Then $\dim W=n-i+1$. Since $$ \dim(U+W)=\dim U+ \dim W -\dim(U\cap W) \leq n, $$ we have $\dim(U\cap W)\geq 1$. Find a unit vector $u \in U\cap W$. Then we have $$ u^t A u \leq \lambda_i, $$ since $u\in W$.
Therefore $\min\limits_{u\in U} u^t Au \leq \lambda_i$ for any choice of $U$ with $\dim U=i$.
Now, we need to achieve that $\lambda_i$ is attained for a choice of $U$. This can be done by choosing $U$ spanned by $\{e_1, \ldots, e_i\}$.