What are all the unramified extensions of $\mathbb{Q}_p$?

Yes, it’s true. Let the residue field of the unramified extension be $\Bbb F_{p^r}=\Bbb F_p(\zeta_{p^r-1})$. The primitive $(p^r-1)$-th root of unity in the residue field can be lifted to characteristic zero — there are any number of ways of seeing this, like Hensel, or the neat trick of lifting $\zeta_{p^r-1}\in\Bbb F_{p^r}$ to any $z$ in your unramified field, and then repeatedly taking $p^r$-th powers. The sequence you get is $p$-adically convergent.

The proof you give is correct. Let me try to exhibit why we should expect such a thing.

The field of complex numbers $\mathbb C$ is the unique finite extension of $\mathbb R$ and that $$ \mathbb R(i) = \mathbb R(\zeta_4) = \mathbb C. $$

Roots of unity are not real except for $\pm 1$. Hence $$ \mathbb R(\zeta_n) = \mathbb R(\zeta_4) =\mathbb C \qquad \text{if $n>2$} $$ and $$ \mathbb R(\zeta_n) = \mathbb R \qquad \text{if $n\leq2$} $$ since there is no more room between $\mathbb R$ and its algebraic closure.

Similar things happen in $\mathbb Q_p$, take $p=5$. By Hensel's lemma roots of unity are not in $\mathbb Q_5$ except for $\pm 1, \pm i$. In particular $$ \mathbb Q_5(\zeta_n) = \mathbb Q_5 \quad \Longleftrightarrow\quad n\mid 4. $$ Now since $\bar{\mathbb Q}_5/\mathbb Q_5$ is an infinite extension there is more room for intermediate extensions but still diferent roots of unity might define the same extension. For instance $\mathbb Q_5(\zeta_{5^2- 1})$ contains $\mathbb Q_5(\zeta_8)$ and $\mathbb Q_5(\zeta_3)$. You can check that $$ \mathbb Q_5(\zeta_{5^2- 1})=\mathbb Q_5(\zeta_8)=\mathbb Q_5(\zeta_3). $$