Is the following claim true? It feels like it should be true, but I don't really know how to show it.
Let $X$ be a Banach space, and $x \in X$ an element of it. Then there exists a functional $\phi \in X^*$ such that $\| \phi \| = 1$ and $\| x \| = | \phi(x) |$.
If I'm not mistaken, it would suffice to say that there exists a sequence $(\phi_k)_{k = 1}^\infty$ of unit functionals for which $| \phi_k (x) | \to \| x \|$, since the unit ball in $X^*$ is compact in the weak topology. However, I don't know how to prove the former result.
EDIT: I forgot to actually define $\psi$ as $\psi(\lambda x) = \lambda$.
My intuition is that I should be able to invoke Hahn-Banach and define a linear function $\psi$ on $\mathbb{C} x \subseteq X$ bounded by the norm $\rho(x) = \| x \|$ on $X$, then extend it from $\mathbb{C} x$ to all of $X$. Is this a correct application of Hahn-Banach?
You don't need completeness of $X$. This is true in normed spaces. Your last idea is a good one:
We may assume $x \neq 0$. Define the functional
$$\varphi: \Bbb{C}x \to \Bbb{C}: \lambda x \mapsto \lambda \Vert x \Vert$$
Then it is easily checked that $\Vert \varphi \Vert =1$ (the inequality $\leq$ is obvious, and then note that $\varphi(x/\Vert x \Vert) = 1$ so also $\Vert \varphi\Vert \geq 1$). By Hahn-Banach, we can extend to a functional $\tilde{\varphi}: X \to \Bbb{C}$ with $\Vert \tilde{\varphi} \Vert =1$ and this is the functional you are looking for.