Show the continuous embedding $ \ell^2 \subseteq c_0. $ 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"?
 A: If $x \in l_2$ then $x \in c_0$.
We have $|x_k| \le \|x\|_2$ for all $ k$, hence $\|x\|_{c_0} = \sup_k |x_k| \le \|x\|_2$.
Just take $c=1$.
A: Question was since edited, rendering this answer obsolete. Here it was:

This is incorrect:

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}

I'm going to assume that you're talking about $\mathcal X$ and $\mathcal Y$ as vector spaces over the same field $\mathbb R$, in which case this is correct.

By definition of $\ell^2$, everything within has a norm:
$$||(x_1,x_2 , \dots )||_{\ell^2}= \sqrt{\sum_{i} x_i^2}$$
The sum converges, which requires that the individual terms of the sum converge to $0$: 
$$\lim_{i\rightarrow \infty} x_i^2 = 0.$$ 
This in turn requires that 
$$\lim_{i \rightarrow \infty} x_i = 0.$$
Note that this condition on $\mathbf x$ is both necessary and sufficient for $\mathbf x \in c_0$, as this is precisely how $c_0$ is defined. 
Hence 
$$\forall \mathbf x \in \ell^2, \quad \mathbf x \in c_0.$$
$$\Rightarrow \quad \ell^2 \subseteq c_0$$
