If $p \geq 2$, then an appropriate version of the Hölder inequality shows that for $U$ bounded, $L^p(U) \hookrightarrow L^2(U)$, hence $W^{1,p}\hookrightarrow W^{1,2} \hookrightarrow L^2$ using the trace inequality. So you have
$$ \|u\|_2 \leq \|u\|_2^{1-1/p}\|u\|_2^{1/p} \lesssim \|u\|_p^{1-1/p} \|u\|^{1/p}_{W^{1,2}} \lesssim \|u\|^{1-1/p}_p \|u\|^{1/p}_{W^{1,p}} $$
which does not require passing through the Gagliardo-Nirenberg inequality. (Though one can argue that the proof of the GN inequality can be recycled to prove the trace theorem, so the two aren't completely unrelated.)
The interesting case is when $p < 2$, in which your estimate follows partially from the generalized Sobolev imbedding theorems (which allows also for trace estimates). See, for example, Theorem 4.12 Case C of Robert Adams' Sobolev Spaces. In particular the numerology requires $p \leq 2 \leq (n-1)p / (n-p)$ which implies
$$ \frac{2n}{n+1} \leq p \leq 2 $$
must hold for the classical trace theorem $W^{1,p}(\Omega) \hookrightarrow L^2(\partial\Omega)$. (If you believe in fractional Sobolev spaces, then you can also "obtain" the above by combining a fractional version of Gagliardo-Niremberg from $W^{1,p}\hookrightarrow H^s$ for some $s > 1/2$ and using the fractional trace theorem $H^s(\Omega)\hookrightarrow L^2(\partial\Omega)$; note that this route naively will require the lower bound on $p$ to be a strict inequality due to the failure of $H^{1/2}(\Omega)\not\hookrightarrow L^2(\partial\Omega)$.) In particular, for dimension $n > 1$ the $L^1$ endpoint is not covered by this theorem.
This failure at $L^1$ is not a problem of the method: it is a genuine failure of the inequality. Consider the function in two dimension $u(x,y) = \left[ (x-1)^2 + y^2\right]^{1/4}$, which is just a translated version of the function $\frac{1}{\sqrt{r}}$. It is not $L^2$ integrable when restricted to the circle, since it has a Logarithmic singularity at $(1,0)$. But $\frac{1}{\sqrt{r}}$ is certainly $L^1_{loc}(\mathbb{R}^2)$, and its derivative which has a $\frac{1}{r^{3/2}}$ singularity is also $L^1_{loc}(\mathbb{R}^2)$. (The same example also works in arbitrary $n \geq 2$; you can either use the same principle via functions like translates of $\frac{1}{r^{(n-1)/2}}$, or just argue through the method of descent.)
As you noted, the authors use single bars $|\cdot |$ for Hilbert space norm, without any subscript. And they use $|\cdot |$ for pointwise vector norm as well, as in (2.11). This made me look for the downvote button, but the journal's site does not have one. Anyway, the first step after the triangle inequality is Cauchy-Schwarz:
$$\|u\cdot \nabla C\|_{L^2(\Omega)}=\left(\int_\Omega |u\cdot \nabla C|^2\right)^{1/2} \le \left(\int_\Omega |u|^4\right)^{1/4}\left(\int_\Omega |\nabla C|^4\right)^{1/4}$$
The second step is the estimate
$$\|u\|_{L^4}\le M\|u\|_{L^2}^{1/2}\|\nabla u\|_{L^2}^{1/2}\tag{1}$$
(and the same for $\nabla C$). At first glance, (1) looks like the Sobolev embedding combined with trivial interpolation between $L^2$ and $L^\infty$ but it's not because $H^1$ does not embed into $L^\infty$ (in nice planar domains it embeds into $L^p$ for $p<\infty$). However, (1) is still true: this is a special case of Gagliardo-Nirenberg interpolation inequality, stated, for example, on page 314 of the book Functional Analysis, Sobolev Spaces and Partial Differential Equations by Haim Brezis. Brezis does not give a proof and refers the reader to Friedman's book which I don't have... but I just noticed it was republished by Dover! Extra special +1 to your question for this.
Best Answer
Both of the inequalities you cite are special cases of a family of interpolation inequalities in $R^n$ that Gagliardo and Nirenberg proved, which have the form $$ \|D^j u\|_{L^p} \le C \|D^m u\|_{L^r}^a \|u\|_{L^q}^{1-a}, $$ for $0\le j<m$ and $\frac jm\le a\le1$, with $$ \frac1p = \frac jn+ a\left(\frac1r-\frac mn\right)+(1-a)\frac1q, $$ except if $1<r<\infty$ and $m-j-n/r$ is an integer then $a=1$ is disallowed. Avner Friedman's PDE book, I believe, contains a full proof.