A proper subspace $W$ of $\mathbb{R^n}$ is closed? Let $W$ be a proper subspace of $\mathbb{R^n}$ equipped with euclidean metric $d$ defined by $d(x,y)= (\sum_{i=1}^{n}(x_i-y_i)^2)^{1/2}$. Let $W$ be a proper subspace of $\mathbb{R^n}$. Then is it closed?
Attempt: Since $W$ is a proper subspace of $\mathbb{R^n} \implies \dim(W) \leq n-1$. Now if $\dim(W)= 0$ then clearly $W$ is closed so we let $\dim(W) = k$, where $1\leq k\leq n-1$. So $W \cong \mathbb{R^k}$ and since $\mathbb{R^k}$ is complete under the above mentioned metric,we can see that any sequence in $\mathbb{R^k}$ converges to a point in $\mathbb{R^k}$ so we can thought of $W$ as $\mathbb{R^k}$ so $W$ is closed.
Is my approach correct? and If I'm wrong, give me hints.

What you did is correct, but I think that it is more natural to do it as follows. Let $\{e_1,\ldots,e_k\}$ be a basis of $W$ and extend it to a basis $\{e_1,\ldots,e_n\}$ of $\Bbb R^n$. Consider the linear map $F\colon\Bbb R^n\longrightarrow\Bbb R^{n-k}$ such that $F(e_{k+1})=(1,0,0,\ldots,0)$, $F(e_{k+2})=(0,1,0,\ldots,0)$ and so on…; besides, $F(w)=0$ for each $w\in W$. Then $W=F^{-1}(\{0\})$ and therefore, since $F$ is continuous and $\{0\}$ is closed, $W$ is closed.
