Proof verification: Every Euclidean space is complete To prove this, I would like to use induction.
For $n=1$ it is easy to prove that $\mathbb{R}$ is complete.
For $n=k$ we assume it is true.
For $n=k+1$, we have to show that $\mathbb {R}^{k+1}$ is complete.
Let $(x_n,y_m) $ be a cauchy seq. in $\mathbb{R}^{k+1}$ and converging to $(x_0,y_0) $ we show that $(x_0,y_0) \in \mathbb {R}^{k+1}$
Now, $\{x_n\}$ is conv. to $x_0$ and $\{x_n\}\subset\mathbb {R}^k $  and $\mathbb{R}^k $ is complete by part 2,so $x_0\in\mathbb{R}^k $
And $\{y_m\}$ converging to $y_0$ and $\{y_m\}\subset\mathbb {R}$ and $\mathbb{R}$ is complete by part 1, so $y_0\in\mathbb{R} $
Hence $(x_0,y_0)\in\mathbb {R}^{k+1} $
Is my proof correct?
 A: No induction needed. Every projection map is distance decreasing so if $(\vec{v})_n$ is Cauchy so are all sequences $(\vec{v}_i)_n$ for $i=1,\ldots,N$, if we're working in $\mathbb{R}^N$.
By completeness of the reals, each of these has a limit $p_i$. And a sequence converges in $\mathbb{R}^N$ iff it converges in every coordinate, so $(\vec{v})_n \rightarrow (p_1,\ldots,p_N)$, as required.
A: No. Your proof is incorrect. As Cornelis said in the comment, your assumption "converging to $(x_0,y_0)$" is logically wrong. 
[Added:] Also, your notation $(x_n,y_m)$ doesn't make sense.
Let $\{(x_n,y_n)\}$ be a cauchy sequence in $\mathbb{R}^{k+1}=\mathbb{R}\times\mathbb{R}^k $. You need to show that it has a limit. 
You need to show that $\{x_n\}$ is Cauchy in $\mathbb{R}$ and $\{y_n\}$ is Cauchy in $\mathbb{R}^k$. Thus by the induction hypothesis, they both have limit, $x_0$ and $y_0$ respectively. Now you show that $(x_n,y_n)\to (x_0,y_0)$ in $\mathbb{R}^{k+1}$. 

[Added:] Here is the induction step for the proof. 
Suppose $\{(x_n,y_n)\}_{n=1}^\infty$ is a Cauchy sequence in $\mathbb{R}\times\mathbb{R}^k$. It follows (this step depends on how you understand Cauchy sequences in $\mathbb{R}^d$ for any positive integer $d$) that $\{x_n\}$ is Cauchy in $\mathbb{R}$ and $\{y_n\}$ is Cauchy in $\mathbb{R}^k$. Using the induction hypothesis, $x_n\to x_0$ in $\mathbb{R}$ and $y_n\to y_0$ in $\mathbb{R}^k$ for some $x_0\in\mathbb{R}$ and $y_0\in\mathbb{R}^k$. Now what you need to show is $(x_n,y_n)\to(x_0,y_0)$ in $\mathbb{R}^{k+1}$.
