I am currently working through this lecture notes and on page 164, there it is said
The space of $\mathcal{D}(\mathbb{R}^n)$ of smooth complex-valued functions with compact support is contained in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$. If $f_k \to f$ in $\mathcal{D}$, then $f_k \to f$ in $\mathcal{S}$, so $\mathcal{D}$ is continuously embedded in $\mathcal{S}$. Furthermore, if $f\in \mathcal{S}$, and $\eta \in C_c^{\infty}(\mathbb{R}^n)$ is a cutoff function with $\eta_k(x) = \eta(x/k)$, then $\eta_k f \to f$ in $\mathcal{S}$ as $k \to \infty$, so $\mathcal{D}$ is dense in $\mathcal{S}$.
I don't understand the arguments in this paragraph, for a subset $\mathcal D$ of $\mathcal S$ to be dense in $\mathcal S$ for every element $s$ of $\mathcal S$ I need to find a sequence in $\mathcal D$ which converges to $s$, but there just stands that $\eta_k f \to f$ in $\mathcal{S}$, but what i need is a sequence in $\mathcal{D}$ not in $\mathcal{S}$, so why does it follow from this that $\mathcal{D}$ is dense in $\mathcal{S}$?