Let $X$ be a normed vector space over $\mathbb K, \mathbb K = \mathbb R$ or $\mathbb K=\mathbb C.$ Let $Y$ be a closed linear subspace of $X$ and $x\in X\backslash Y.$ Set $Z=\{y+\alpha x;\;y\in Y,\;\alpha\in\mathbb K\}.$ Show that $Z$ is closed.
Hint: You can use without proof that every bounded sequence in $\mathbb K$ has a convergent subsequence and every unbounded sequence $(\alpha_n)$ a subsequence with $1/|\alpha_{n_j}| \rightarrow 0$ as $j\rightarrow\infty.$
Let $z\in \overline Z$, then there exists $z_n\in Z$ such that $z_n\to z$.
$z_n=y_n+\alpha_n x$ with $y_n\in Y$ and $\alpha_n\in \Bbb K$.
If $\alpha_n$ is bounded, then it admit a convergent subsequence $\alpha_{n_j}$, and denote $\alpha$ its limit. We get $y_{n_j}=z_{n_j}-\alpha_{n_j}x$ which is convergence, as $Y$ is closed , we have $y_{n_j}\to y=z-\alpha x\in Y$, hence $z=y+\alpha x\in Z$.
If $\alpha_n$ is note bounded then it admit a subsequence $\alpha_{n_j}$ such that $|\alpha_{n_j}|\to \infty$, hence $\beta_{n_j}=\dfrac{1}{\alpha_{n_j}}\to 0$. but $\beta_{n_j}z_{n_j}=\beta_{n_j}y_{n_j}+x$, this implies that $\beta_{n_j}y_{n_j}=\beta_{n_j}z_{n_j}-x\to x$ since $\beta_{n_j}y_{n_j}\in Y$ and $Y$ closed , we get $x\in Y$ (impossible).
Conclusion $z\in Z$, and $Z$ is closed.