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The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.