It seems this can be done "by hand", using much less than what goes into the full proof of Chebotarev. (But it's the same circle of ideas.) It is known that for any number field $K/\mathbf{Q}$, we have $\zeta_K(s) \sim \frac{\kappa}{s-1}$ as $s\downarrow 1$, for some $\kappa > 0$. Hence, $\log \zeta_K(s) = \log \frac{1}{s-1} + O(1)$, as $s\downarrow 1$. On the other hand, starting from the Euler product representation $\zeta_K(s) = \prod_{\mathfrak{p}} (1-|\mathfrak{p}|^{-s})^{-1}$, we also see that $\log\zeta_K(s) = \sum_{\mathfrak{p}} \frac{1}{|\mathfrak{p}|^s} + O(1) = \sum_{p} \frac{n(p)}{p^s} + O(1)$, where $n(p)$ is the number of degree one prime ideals $\mathfrak{p}$ of $\mathcal{O}_K$ that lie above the rational prime $p$. (Note that I'm using the somewhat nonstandard notation $|\cdot|$ for the norm map on ideals.)
In particular, when $K=\mathbf{Q}$, we get that $\sum_{p} \frac{1}{p^s} = \log\frac{1}{s-1} + O(1)$ as $s\downarrow 1$.
Now suppose that $k$ is a number field with the property that all but finitely many rational primes $p$ have a degree $1$ prime $\mathfrak{p}$ of $\mathcal{O}_k$ lying above them. Applying the result of the first paragraph to this $k$, we see that
$$ \log \frac{1}{s-1} \sim \log \zeta_k(s) = \sum_{p} \frac{n(p)}{p^s} + O(1)\geq \sum_{p} \frac{1}{p^s} + O(1) \sim \log \frac{1}{s-1}. $$
Hence,
$$ \sum_{p:~n(p)>1}\frac{n(p)-1}{p^s} = o\left(\log\frac{1}{s-1}\right), $$
as $s\downarrow 1$. Consequently, $$ \sum_{p:~n(p)>1}\frac{1}{p^s} = o\left(\log\frac{1}{s-1}\right). $$
That is, the (Dirichlet) density of primes $p$ having more than one degree one prime ideal above them must be zero. But unless $[k:\mathbf{Q}] = 1$, this contradicts that a positive proportion of rational primes split completely in $k$. (It is in this final step that $n > 1$ is important.)
It might seem that I smuggled in Chebotarev to say that a positive proportion of rational primes split in $k$. But no, this follows again from the result in the first paragraph, now with $K$ taken as the Galois closure of $k/\mathbf{Q}$.
