Show the continuous embedding $$ \ell^2 \subseteq c_0. $$
I couldn't find a question that showed this "simple" proof, and I'm having trouble doing it myself.
For normed spaces $\mathcal{X},\mathcal{Y}$, let $x \in \mathcal{X}$. Then \begin{equation} \mathcal{X} \subseteq \mathcal{Y} \iff \exists c > 0 \;:\;\Vert x \Vert_{\mathcal{Y}} \le c \Vert x \Vert_{\mathcal{X}} \quad \forall x\in\mathcal{X}. \end{equation}
We say normed space $\mathcal{X}$ continuously embeds into normed space $\mathcal{Y}$.
The sequence spaces $\ell^2,c_0$ have norms $$\Vert x \Vert_2 = \left( \sum_{n} |x_n|^2 \right)^{1/2}$$ and $$\Vert x \Vert_{c_0} = \sup_{n} |x_n|,$$ respectively.
We say $x \in \ell^2 \iff \Vert x \Vert_2 < \infty$ and $x \in c_0 \iff \lim_{n\to\infty} x_n = 0$.
My attempt:
Suppose $x \in \ell^2$. Then $\lim_{n \to \infty} |x_n| = 0$. Hence, there exists $k$ such that $x_k = \sup_{n} |x_n|$. Therefore,
\begin{align} \mathrm{LHS} & = \sup_{n} |x_n| \\ & = |x_k| \\ & \le \sum_{n} |x_n| \text{ since $|x_n| \ge 0$ for $n\ne k$} \\ & \le c \left( \sum_{n} |x_n|^2 \right)^{1/2} \end{align} since it is true that for all $u,v\in\mathbb{R}$, $u \ge 0$, $v \ge 0$, there exists $c \ge 0$ such that $u \le cv$. Hence, $\ell^2 \subseteq c_0$.
Is the property of real numbers that I used enough to justify the ending of my "proof"?