Proof that a constraints set is closed

Solution 1:

Your set is the intersection of the sets$$\{\mathbf x\in\Bbb R^n\mid\langle\mathbf c,\mathbf x\rangle=0\}\tag1$$and$$\{\mathbf x\in\Bbb R^n\mid\psi(\mathbf x)\leqslant0\}.\tag2$$All you need to do then is to prove that both $(1)$ and $(2)$ are closed sets. But if $\varphi\colon\Bbb R^n\longrightarrow\Bbb R$ is the function defined by $\varphi(\mathbf x)=\langle\mathbf c,\mathbf x\rangle$, then$$(1)=\varphi^{-1}\bigl(\{0\}\bigr)\quad\text{and}\quad(2)=\psi^{-1}\bigl(]-\infty,0]\bigr).$$So, since $\varphi$ and $psi$ are continuous and $\{0\}$ and $]-\infty,0]$ are closed subsets of $\Bbb R$, $(1)$ and $(2)$ are indeed continuous.

Concerning your attempt, I see no reason why your equalities between sets would hold.