the problem above is in Conway's Functional Analysis Page $93$.
it seems to be an application of closed graph theorem
if the inequality were posed the other way it could be much easier
but it is hard to resolve that in this setting
anybody who knows some hint for this?
As mentioned in the comments, consider the restriction map: $$ \Phi: \mathfrak{X}\to C(E) $$ It is not difficult to show that $\Phi$ is linear and bounded. In fact, $\Phi$ is surjective by the assumptions in the question. Therefore, $\Phi$ admits a factorization by the first isomorphism theorem: $$ \mathfrak{X}\xrightarrow{P}\frac{\mathfrak{X}}{\ker\Phi} \xrightarrow{\overline{\Phi}} C(E) $$ Here, $P$ is the projection onto the quotient, and $\overline{\Phi}$ is an isomorphism of Banach spaces. Therefore, one can call upon the Bounded Inverse Theorem, which gives that $\overline{\Phi}^{-1}:C(E)\to \mathfrak{X}/\ker\Phi$ is bounded.
We're pretty much done, because we have a bounded map in the right direction. Now, choose any element $g\in C(E)$, and consider the coset $[f] = \overline{\Phi}^{-1}(g)$. Since $\overline{\Phi}^{-1}$ is bounded, we have that: $$ \left\lVert [f] \right\rVert \leq \lVert \overline{\Phi}^{-1}\rVert \left\lVert g\right\rVert $$ But, by the definition of the quotient space norm, for every $\varepsilon > 1$, one can find an actual element $f'\in \mathfrak{X}$ such that: $$ \left\lVert f' \right\rVert \leq \varepsilon\left\lVert [f] \right\rVert = \varepsilon \inf_{h\in\ker\Phi} \left\lVert f+h \right\rVert $$ Then, $f'$ is a solution, and satisfies the inequality: $$ \left\lVert f' \right\rVert \leq \varepsilon \left\lVert [f] \right\rVert \leq \varepsilon \lVert \overline{\Phi}^{-1}\rVert \left\lVert g\right\rVert $$