My code can't be uploaded because it doesn't work with the websites coding, but here is a pdf of my LaTex code. My question is, is this a proper proof? It feels as if I'm missing something important. Perhaps defining $m$ to be the max of the $N_1,\ldots,N_n$? 
Edit:
Let $\{( x_{n,1},x_{n,2},\ldots,x_{n,m})\}_{n\in\mathbb{N}}$ be a Cauchy sequence in $\mathbb{R}^m$. Let $\epsilon>0$. Thus we have that: \begin{align*} \forall \thickspace \{( x_{n,1},x_{n,2},\ldots,x_{n,m})\}_{n\in\mathbb{N}},\thickspace \rho(x_{n,k},x_{m,k}) < \epsilon. \end{align*} Then each $ \{( x_{n,1},x_{n,2},\ldots,x_{n,m})\}_{n\in\mathbb{N}\}}$ is a Cauchy sequence in $\mathbb{R}$. Since each Cauchy sequence is in $\mathbb{R}$, then by completeness of $\mathbb{R}$, we have that: \begin{align*} \exists \thinspace y_1,y_2,\ldots,y_m \in \mathbb{R} \thickspace \mbox{and} \thickspace N_1,N_2,\ldots,N_m \in \mathbb{N}\thickspace (\mbox{depending on}\thickspace \epsilon) \thickspace \mbox{such that} \end{align*} \begin{align*} &\forall \thinspace k\in \mathbb{N} : 1\leq k \leq m : \forall \thinspace n_k > N_k : \rho(x_{n,k},y_k) < \frac{\epsilon}{\sqrt{m}} \end{align*} Following from above, we have that by definition of $\rho$ and the fact that $\rho(x,y) \geq 0$, that: \begin{align*} \sqrt{\sum_{k=1}^m|x_{n,k}-y_k|^2} < \frac{\epsilon}{\sqrt{m}}& \implies \sum_{k=1}^m|x_{n,k}-y_k|^2 < \frac{\epsilon^2}{m} \\ &\implies |x_{n,k}-y_k|^2 < \frac{\epsilon^2}{m} \\ &\implies |x_{n,k}-y_k| < \frac{\epsilon}{\sqrt{m}}. \end{align*} But we know that $|x_{n,k}-y_k|$ is just a sequence of real numbers and we know that $(\mathbb{R},|\cdot|)$ is a complete metric space, so $|x_{n,k}-y_k|$ converges in $\mathbb{R}$. More specifically, each $x_{n,k} \rightarrow y_k$ for $k\in\mathbb{N}$. This is true due to: $\begin{align} \lim_{n\rightarrow \infty} (x_{n,1},x_{n,2},\ldots,x_{n,k}) = (\lim_{n\rightarrow\infty} x_{n,1},\lim_{n\rightarrow\infty} x_{n,2},\ldots,\lim_{n\rightarrow\infty}x_{n,m}) = (y_1,y_2,\ldots,y_m)\in\mathbb{R}. \end{align}$
One way to see that there’s a problem is to notice that you never actually used the crucial hypothesis that the original sequence is Cauchy. I’ll give you an outline for the proof; see if you can fill in the details.
You have the original Cauchy sequence $\langle x_n:n\in\Bbb N\rangle$, where $x_n=\langle x_{n,k}:k\in\Bbb N\rangle$. Use the fact that it’s Cauchy to show that each of the coordinate sequences $\langle x_{n,k}:n\in\Bbb N\rangle$ for $k=1,\ldots,m$ is a Cauchy sequence in $\Bbb R$.
Use the fact that $\Bbb R$ is complete to show that for each $k\in\{1,\ldots,m\}$ there is a $y_k\in\Bbb R$ such that $\langle x_{n,k}:n\in\Bbb N\rangle\to y_k$.
Now show that $\langle x_n:n\in\Bbb N\rangle\to\langle y_1,\ldots,y_m\rangle$ in $\Bbb R^m$.