My Try
(a)$K\subset M_n(\mathbb R)\sim \mathbb R^{n^2}$ compact iff $K$ is closed and bounded. $A\in K$ has entries from the bounded set. So, eigenvalues are bounded.
(b) $tr(A)=1$ is a hyperplane in $M_n(\mathbb R)\sim \mathbb R^{n^2}$. $A=A^t$ is a subspace of this hyperplance with dimension $n(n+1)/2$. $A$ is positive semi-definite. So eigen values of $A$ are non negative. I am not conclude anything.
(c) Let $\{\mathcal U_\alpha\}_{\alpha \in \Lambda}$ be an open cover for $K$. I wanted to find a finite subcover for this OR find an open cover which has no finite sub cover.
In (a), you are missing an argument showing why bounded entries implies bounded eigenvalues. One way to show this is by considering the characteristic polynomial, whose coefficients are bounded (as they are sums of determinants of the matrix entries).
In (b), you can show it is compact by using the non-negativity of the eigenvalues together with the trace condition to exhibit an explicit upper bound on the Frobenius norm (i.e. the sum of the squares of all matrix entries).
In (c), we can see that $K$ is non-compact by the following simple (yet indirect) argument. Suppose for contradiction that $K$ is compact, and consider the function $K\to\mathbb R$ given by $f\mapsto \sup f$. This is continuous, basically by definition of the metric you are given on $K$. However, $f(K)=[0,\infty)$ which can easily be seen by considering a sequence of probability distributions concentrating on a single point. In particular, $f(K)$ is non-compact, contradicting the theorem stating that the continuous image of a compact set is compact.