Let $X$ be a reflexive Banach space, and $X'$ it's dual space. Let $f \in X'$, $f \neq 0$ and $\alpha\in\mathbb{F}$. The hyperplane in $X$ defined by $f$ and $\alpha$ is the set $$H:=\{x\in X:f(x)=\alpha\}.$$ Show that there exists a $z\in H$ such that $\inf_{h\in H}\lVert h\rVert=\lVert z\rVert$.
Since we know that $X$ is reflexive I tried using the fact that for every $f \in X'$ there exists an $x \in X$ with $\lVert x \rVert = 1$, such that $f(x) = \lVert f \rVert$, but I still can't prove the fact that $x$ is an element of $H$.
There is no reason why $f(x)$ should be equal to $\alpha$, and as a consequence, no reason why $x$ should belong to $H$.
Still, your idea is a good starting point. What we can do is find a vector $z$ colinear to $x$ and belonging to $H$. If $z = \lambda x \in H$, then $f(z) = \lambda f(x) = \lambda \|f\| = \alpha$, so the right choice is
$$z = \frac{\alpha}{\|f\|} x.$$
By construction, $z \in H$, and in addition, $\|z\| = \frac{|\alpha| \|x\|}{\|f\|} = \frac{|\alpha|}{\|f\|}$.
All is left is to check that $z$ realizes the minimum to the distance to $0$, or, in other words, that $\|h\| \geq \frac{|\alpha|}{\|f\|}$ whenever $h \in H$. But, for any $h \in H$,
$$\alpha = f(h)$$
$$|\alpha| = |f(h)| \leq \|f\| \|h\|,$$
and thus $\|h\| \geq \frac{|\alpha|}{\|f\|}$, as desired.