Convexity of set $\lambda_{\min}(M) \geq a$ in the space of symmetric matrices

This comes down to the fact that two choices are better than one. First recall: $\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\text{trace}\big(\mathbf x\mathbf x^T M\big)=\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\mathbf x^T M\mathbf x = \lambda_\min(M)$

Thus for arbitrary $M, M' \in C$ and $p \in [0,1]$ and $q:=1-p$

$\lambda_\min\big(pM +qM'\big)$
$= \min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\text{trace}\Big(\mathbf x\mathbf x^T \big(pM +qM'\big)\Big)$
$=\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf x\mathbf x^T \big(qM'\big)\Big)\Big\}$
$=\min_{\mathbf x, \mathbf y: \Vert \mathbf x\Vert_2=1 \& \Vert \mathbf y\Vert_2=1 \&\mathbf x:=\mathbf y}: \Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf y\mathbf y^T \big(qM'\big)\Big)\Big\}$
$\geq \min_{\mathbf x, \mathbf y: \Vert \mathbf x\Vert_2=1 \& \Vert \mathbf y\Vert_2=1 }: \Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf y\mathbf y^T \big(qM'\big)\Big)\Big\}$
$= p\cdot \min_{\mathbf x:\Vert \mathbf x\Vert_2=1 }: \text{trace}\Big(\mathbf x\mathbf x^T M\Big)+q\cdot \min_{\mathbf y:\Vert \mathbf y\Vert_2=1 }:\text{trace}\Big(\mathbf y\mathbf y^T \big(M'\big)\Big)$
$\geq p\cdot a +q\cdot a$
$=a$
$\implies \big(pM +qM'\big) \in C$