Let $(X,\|\cdot\|)$ be a normed space and $Y$ be a closed subspace of $X$, such that $Y\neq X.$ If $x_0\in X\setminus Y,$ prove that the subspace $<Y\cup \{x_0\}>$ is closed in $X$.
Attempt. Let $x\in X$ be in the closure of $<Y\cup \{x_0\}>$, so $y_n+\lambda_nx_0\rightarrow x$ for some $(y_n)$ in $Y$ and $(\lambda_n)$ in $\mathbb{R}.$
Claim 1. $(\lambda_n)$ is Cauchy. Then $\lambda_n$ converges to some $\lambda$ and $$y_n=y_n+\lambda_nx_0-\lambda_nx_0\rightarrow x-\lambda x_0$$ so $x=(x-\lambda x_0)+\lambda x_0\in <Y\cup \{x_0\}>$.
or
Claim 2. $(\lambda_n)$ is bounded. Then some subsequence $(\lambda_{k_n})$ converges to some $\lambda$ and $$y_{k_n}=y_{k_n}+\lambda_{k_n}x_0-\lambda_{k_n}x_0\rightarrow x-\lambda x_0$$ so $x=(x-\lambda x_0)+\lambda x_0\in <Y\cup \{x_0\}>$.
I am not sure which one of the above claims is correct (if so) and I can't prove either. Any help would be appreciated. Thanks in advance.