Correctness of convergence proof

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The problem is this :

Given a sequence $\{x_k\}_{k = 1}^{\infty} \subset \mathbb{R}^n, x_k, x \neq 0$, define $y_k = \dfrac{x_k}{\|x_k\|}$. If $x_k \to x \in \mathbb{R}^n$ then prove that $y_k \to \dfrac{x}{\|x\|}$.

My proof is as follows: $x_k \to x \implies c_k := \dfrac{1}{\|x_k\|} \to \dfrac{1}{\|x\|}$. Now $x_k \to x$, so $y_k = c_kx_k \to \dfrac{x}{\|x\|}$ as $c_k \to \|x\|$.

This uses the fact that if $c_k \to c \in \mathbb{R}, x_k \to x \in \mathbb{R}^n \implies c_kx_k \to cx$, whose proof proceeds by using the fact that $c_k x_{k, i} \to cx_i$. But isn't this approach somewhat flawed, in that $c_k$, in our case, depends on all $x_{k, j}\ (j = 1, \ldots n)$.

Is it allowed to separate and work on individual convergences, instead of showing that the entire thing converges as a whole. I'm having kind of a conceptual block here.

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If $c_{k}\rightarrow c$, for some $M>0$, $|c_{k}|\leq M$ for all $k=1,2,...$, so \begin{align*} |c_{k}x_{k}-cx|&=|c_{k}x_{k}-c_{k}x+c_{k}x-cx|\\ &\leq|c_{k}|\cdot|x_{k}-x|+|x|\cdot|c_{k}-c|\\ &\leq M|x_{k}-x|+|x|\cdot|c_{k}-c|, \end{align*} now $M|x_{k}-x|+|x|\cdot|c_{k}-c|\rightarrow 0$ as $k\rightarrow\infty$, now use Squeeze Theorem to conclude that $c_{k}x_{k}\rightarrow cx$.

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Hint: Use the fact that every norm in $\mathbf R^n$ is continuous, that is, $x_k \to x$ implies $\| x_k\| \to \|x\|$.

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You need to assume that $x \neq 0$.

Then \begin{eqnarray} \| {x \over \|x\|} - { x_n \over \|x_n\| } \| &=& \| { \|x_n\| x - \|x\| x_n \over \|x\| \|x_n\| }\| \\ &\le& \| { \|x_n\| x - \|x_n\| x_n \over \|x\| \|x_n\| }\| + \| { \|x_n\| x_n - \|x\| x_n \over \|x\| \|x_n\| }\| \\ &\le & {1 \over \|x\| } \|x-x_n\| + {| \|x\|-\|x_n\| |\over \|x\| |} \\ &\le& {2 \over \|x\|} \|x-x_n\| \end{eqnarray} where I used $| \|x\|-\|x_n\| | \le \|x-x_n\|$ on the last line.