Proof.

1. (1)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $E\cup F$ then there exists $J,J'\suf I$ finite such that $\cups{}{j\in J}U_{j}\supf E$ and $\cups{}{j\in J'}U_{j}\supf F$ hence $\cups{}{j\in J\cup J'}U_{j}\supf E\cup F$.

2. (2)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $G$. Since $G$ is closed, $\curl{U_i}_{i\in I}\cup \curl{G^c}$ is an open cover of $E$. In which case, there exists $J\suf I$ finite such that $\cups{}{j\in J}U_{j}\cup G^{c}\supf E$, so $\cups{}{j\in J}U_{j}\supf G$.

3. (3)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $f(E)$, then $\curl{f\inv (U_i)}_{i\in I}$ is an open cover of $E$. Thus there exists $J\suf I$ finite such that $\cups{}{j\in J}f\inv(U_{j})\supf E$, so $\cups{}{j\in J}U_{j}\supf f(E)$.

$\square$