Convergence of vectors in $\mathbb{R}^m$ proof - help understanding a step? The theorem is as follows:
A sequence of vectors in $\mathbb{R}^m $ ($m \in \Bbb Z_+$) converges iff all $m$ sequences of its components converge in $\mathbb{R}^1$.
I can follow the (if) direction of the proof, but am struggling with the only if part.
Here it is:
Suppose $ \{\mathbf{x}_n\}_{n=1}^\infty$ converges to $\mathbf{x}$. 
Choose $\varepsilon < 0$, then there exists an integer $N$ s.t $\forall  n_{\in \Bbb Z} \geq N$
$$ ||\mathbf{x}_n - \mathbf{x^*} || < \varepsilon  $$
But then, for $n \geq N$ for each component $i \in \{1, ... , m\}$:
$$|x_{in} - x_i ^*| \leq \sqrt{(x_{1n}-x_1^*)^2 + ... + (x_{mn}-x_{m}^*)^2}$$
$$ = ||\mathbf{x}_n - \mathbf{x^*}|| < \varepsilon$$
I just wanted to clarify that second last step - with the inequality. I have a feeling this is because of the triangle inequality applied in $ \mathbb{R}^m$, and then the final equality is just using the definition of the $\ell_2$-norm?
To be honest, as I was writing this question out, I feel like I started to actually understand it, but since it's written it'd be great if I could get confirmation.
Thanks.
 A: The inequality arises due to a much simpler reason:
$$
|x_{in} - x^\ast_{i}|^2 = \color{red}{(x_{in} - x^\ast_{i})^2} \leq \color{blue}{(x_{1n} - x^\ast_{1})^2} + \cdots + \color{red}{(x_{in} - x^\ast_{i})^2} + \cdots + \color{blue}{(x_{mn} - x^\ast_{m})^2}
$$ The LHS is less than the RHS simply because the RHS includes the LHS $\color{red}{\text{red }(x_{in} - x^\ast_{i})^2}$ term along with possibly other, non-negative $\color{blue}{\text{blue}}$ terms like $\color{blue}{(x_{1n} - x^\ast_{1})^2}, \color{blue}{(x_{mn} - x^\ast_{m})^2}, \text{etc}$.
Thus we are just using the elementary fact that if $b \in \Bbb R$ and $\boxed{b \geq 0}$ then $\forall a \in \Bbb R$ $$a \leq a + b = b + a$$ again and again above.
Now taking the root on both sides gives
$$
|x_{in} - x^\ast_{i}| \leq \sqrt{(x_{1n} - x^\ast_{1})^2 + \cdots + (x_{in} - x^\ast_{i})^2 + \cdots (x_{mn} - x^\ast_{m})^2} = ||\mathbf{x}_n - \mathbf{x}^\ast||
$$
A: Regarding the inequality
$$|x_{in} - x_i^*| \le \sqrt{(x_{1n} - x_1^*)^2 + (x_{mn} - x^*_m)^2},$$
try squaring both sides. (No triangle inequality needed.)
