Show $A = \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact set The problem
Let $\left( E, \langle \cdot, \cdot \rangle \right)$ be a euclidean space of dimension $n$. Let $K$ be a compact subset of $E$ containing a basis $e = (e_1, ..., e_n)$ of $E$.
We denote by $S^+(E)$ the set of self-adjoint endomorphisms of $E$ with non-negative eigenvalues.
Let I want to show that $A := \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact subset of $\mathcal{L}(E)$.
I have already proved $A$ is a closed subset of $\mathcal{L}(E)$, using sequential characterisation and without using the compactness of $K$. So, what remains to prove is the boundedness of $A$. 
What I have tried
Let $u \in A$. The spectral theorem gives us an orthonormal basis $f = (f_1, ..., f_n)$ composed of eigenvectors of $u$, with eigenvalues $\lambda_1, ..., \lambda_n$. 
We decompose each $e_i = a_{1i}f_1 + ... + a_{ni}f_n$ (so that $M := (a_{ij}) = \textrm{Mat}_{f}(e)$). As $f$ is orthonormal and $e_i \in K$, one has $\langle e_i, u(e_i) \rangle = a_{1i}^2\lambda_1 + ... + a_{ni}^2\lambda_n \leq 1$. Denoting $\Lambda := (\lambda_1, ..., \lambda_n)^T$ and $N := (a_{ij}^2)$ we can claim that $N \Lambda$ is bounded. 
My goal now is to show that $\Lambda$ is bounded by something which does not depend on $u$. Indeed, the spectral radius is a norm on $S(E)$.
But there is one main several obstacle: I do not manage to get rid of $N$, which depends on $u$, because the $a_{ij}$ do. Furthermore, I suspect the compactness of $K$ is useful here, but I cannot figure out any link with it.
 A: It is a little funny the way the problem is written. Here is what you need to show. Consider the matrices $$M_n^+(\mathbf{R}) = \{P \in \mathbf{R}^{n \times n} : P = P^T, P \succeq 0\}$$ 
These are canonically identified with the set $S^+(E)$ of self-adjoint, endomorphisms you give before (viz. the spectral theorem). 
Consequently, you only need to show that 
$$
A = \left\{U \in M_n^+(\mathbf{R}) : \sup_{x \in K} x^T U x \leq 1\right\}$$ 
is a compact set. Closure is simple: if $U_n \to U$, $U_n \in A$, then 
$x^TU_n x \leq 1$ for all $x \in K$. Moreover the sequence $\{x^TU_n x\}$ is bounded since $x$ are bounded and $U_n$ converges. Consequently, we see that $x^T U x \leq 1$, and therefore $A$ is closed. 
For bounded, it is quite easy. The operator norm is always achieved in finite dimensions. So let $x$ be a unit vector, achieving the operator norm for $U^{1/2}$, with $U \in A$. Then we may write $x = \sum_i c_i e_i$, where $e_i$ denote the basis contained in $K$. Then 
$$\|U^{1/2}\| = \|U^{1/2} x\| \leq \sum_i |c_i| \|U^{1/2} e_i\| = \sum_{i=1}^n |c_i| \leq B.
$$
The existence of $B$ follows by compactness.
Moreover, $\|U\| \leq \|U^{1/2}\|^2 \leq B^2$. So $A$ is bounded. 
