Why is the Minimum in the Min-Max Principle for Self-Adjoint Operators attained? Let's consider a self-adjoint operator $A$ (not necessarily bounded) on a Hilbert space which is bounded from below, with domain $D$ and whose resolvent is compact. Then, the spectrum consists solely of isolated eigenvalues which are given (in increasing order) by the min-max principle:
\begin{equation}
\lambda_k = \min_{\substack{V \subset D\\ \dim V  = k}} \max_{\substack{x \in V \\ x \neq 0}} \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}, \ k \in \mathbb{N}.
\end{equation}
The proof I know shows $\lambda_k \geq \min \max \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}$ and $\lambda_k \leq \min \max \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle}$ by using a orthonormal basis of eigenvectors.
As seen here: Why is the Maximum in the Min-Max Principle for Self-Adjoint Operators attained?, we know that the maximum is attained since the unit sphere is compact in a finite dimensional vector space.
But why is the minimum also attained?
 A: In the finite dimensional case, we get a similar argument to work: the max is continuous (as a function of $V$), and the space of $k$-dimensional vector spaces is compact, so the minimum is reached. 
In the infinite dimensional case, the main difference is that $\mathcal{G} (k, \infty)$ is no longer compact. It is still a metric space, and the application $F: V \mapsto \max_{x \in V, \|x\|=1} \langle x, Ax \rangle$ is still continuous. I am not sure about the best way to make this line of reasoning work in this case, although I strongly suspect that $F$ is proper (so its infimum is a minimum). One would have to use the fact that the resolvent is compact, though.
At worse, one can use the spectral decomposition to show that the minimum is reached. Take the first $k$ eigenvalues, take corresponding unit eigenvectors $(e_i)_{1 \leq i \leq k}$ (choosing them orthogonal if necessary), and put $V = Span((e_i))$. But this can only be done once we know the spectral decomposition, and I would prefer a more elementary (and geometrical) proof.
Now, I'll sketch a proof of the geometrical argument. There are a few holes, which should not be too hard to fill in.

1) Let $\mathcal{G} (k,n)$ be the space of $k$-dimensional vector subspaces in $\mathbb{C}^n$ (this is also called a Grassmannian). Let $(V_j)$ be a sequence in $\mathcal{G} (k,n)$, and $V \in \mathcal{G} (k,n)$. We say that $\lim_j V_j = V$ is there exists a sequence $(e_1^{(j)}, \ldots, e_k^{(j)})$ of orthonormal bases of $V_j$, and an orthonormal basis $(e_1, \ldots, e_k)$ of $V$, such that $\lim_j e_i^{(j)} = e_i$ for all $1 \leq i \leq k$.
This topology is metrizable. For instance, you can define:
$$d (V_1, V_2) := \inf \max_{1 \leq i \leq k} \{\|e_i-f_i\|\},$$
where the infimum runs over all orthonormal bases $(e_i)$ of $V_1$ and $(f_i)$ of $V_2$. It's easy to check that $d$ is symmetric, non-negative and satisfies the triangle inequality, and that $d(V,V) =0$ for all $V$. Proving that $d(V_1, V_2) >0$ if $V_1 \neq V_2$ is harder.
Let $V_1 \neq V_2$ be in $\mathcal{G} (k,n)$. Let $u_1 \in V_1 \setminus V_2$ with unit length. Then there exists $\varepsilon >0$ such that $| \langle u_1, v_2 \rangle | \leq 1-\varepsilon$ for all unit $v_2 \in V_2$ (using the compactness of $\mathbb{S}_{n-1}\cap V_2$). Let $(e_i)$ and $(f_i)$ be orthonormal bases of $V_1$ and $V_2$ respectively. Write $u_1 = \sum_i \lambda_i e_i$, with $\sum_i |\lambda_i|^2 = 1$. Then:
$$2\varepsilon \leq 2-2 \mathfrak{Re} \langle u_1, \sum_i \lambda_i f_i \rangle = \left\| u_1 - \sum_i \lambda_i f_i\right\| \leq \sum_i |\lambda_i| \|e_i-f_i\| \leq \sum_i |\lambda_i| d(V_1,V_2) \leq k d(V_1,V_2)$$
Hence, $d(V_1, V_2) >0$. Thus, $d$ is a distance.

2) In addition, $\mathcal{G} (k,n)$ is compact. Let $\mathcal{E} (k,n)$ be the space of all $k$-uplets of orthonormal vectors (or the space of orthonormal $k \times n$ matrices). Then $\mathcal{E} (k,n)$ is closed and bounded in the space of $k \times n$ matrices, so compact, and there is an application $\varphi: \mathcal{E} (k,n) \to \mathcal{G} (k,n)$, defined by:
$$\varphi (e_1, \ldots, e_k) = Span (e_1, \ldots, e_k).$$
This application is continuous, and even $1$-Lipschitz (it follows easily from the definition of the distance $d$).
Hence, the image of $\varphi$ is compact. But $\varphi$ is surjective, so $\mathcal{G} (k,n)$ is compact.

3) Let us define, for $V \in \mathcal{G} (k,n)$:
$$F(V) := \max_{\substack{x \in V \\ x \neq 0}} \frac{\langle x, Ax\rangle}{\langle x, x \rangle} = \max_{\substack{x \in V \\ \|x\|=1}} \langle x, Ax \rangle$$
Let $V_1$, $V_2$ be in $V$. Let $(e_i)$, $(f_i)$ be bases of $V_1$ and $V_2$ respectively. Let $u := \sum_i \lambda_i e_i \in \mathbb{S}_{n-1} \cap V_1$. Then:
$$\langle \sum_i \lambda_i f_i, A\sum_i \lambda_i f_i \rangle \leq \langle u, Au \rangle+\left| \langle \sum_i \lambda_i (f_i-e_i), Au \rangle \right|+\left| \langle \sum_i \lambda_i (f_i-e_i), A\sum_i \lambda_i f_i \rangle \right| \leq \langle u, Au \rangle + 2 \|A\| \left\| \sum_i \lambda_i (f_i-e_i) \right\| \leq \langle u, Au \rangle + 2k \|A\| \max_i \|f_i-e_i \|$$
Since this is true for all $(\lambda_i)$, we get, for all $v \in V_2$:
$$\langle v, Av \rangle \leq \max_{\substack{u \in V_1 \\ \|u\|=1}} \langle u, Au \rangle+2k\|A\| \max_i \|f_i-e_i \|.$$
Since this holds for all bases $(e_i)$ and $(f_i)$, we can choose them as close as possible, whence:
$$\langle v, Av \rangle \leq \max_{\substack{u \in V_1 \\ \|u\|=1}} \langle u, Au \rangle+2k\|A\| d(V_1, V_2).$$
Finally, taking the maximum over all unit $v \in V_2$, we get:
$$F(V_2) \leq F(V_1)+2k\|A\| d(V_1, V_2).$$
If we exchange $V_1$ and $V_2$, we get $F(V_1) \leq F(V_2)+2k\|A\| d(V_1, V_2)$. Hence, $F$ is Lipschitz, and thus continuous.
Since $\mathcal{G} (k,n)$ is compact and $F$ is continuous on $\mathcal{G} (k,n)$, the function $F$ reaches its minimum.
