As the title implies, I have a couple of problems that I worked on for an introductory Analysis course and want to make sure that they're valid.
My attempt: $E$ is connected if the only subset of $E$ that are both open and closed are $E$ and the empty set. We are given that $\{x_0\}$ is open; any sequence in this set consists of only the point $x_0$ and therefore converges to $x_0$. A set is closed if every sequence contained in the set converges to a point also in the set, so now we have an open and closed set $\{x_0\}$; therefore $E$ is not connected. $\square$
My attempt: Pick a sequence of points $\{x_n\}$ contained in $S$. We want to show that it has a convergent subsequence. Since $S$ is a closed subset of a complete space, then it $S$ is also complete. Therefore, if $\{x_n\}$ is Cauchy, then it is convergent/has a convergent subsequence. Assume $\{x_n\}$ is not Cauchy. Then, by the given property for $E$, $\{x_n\}$ is not bounded and therefore cannot be contained in $S$. It follows that every sequence in $S$ is Cauchy and has a convergent subsequence, so $S$ is compact. $\square$
Mainly I am not entirely confident in my showing that $\{x_0\}$ is closed (for the first problem), and the usage of the property of $E$ (in the second problem), but I would of course appreciate feedback on the proofs overall. Thank you very much!
You should have posted two distinct questions, but I will answer.
First question: Your proof is correct, but it can be made simpler. The set $\{x_0\}$ is closed because$$E\setminus\{x_0\}=\bigcup_{x\neq x_0}B_{d(x,x_0)}(x)$$and therefore $E\setminus\{x_0\}$ is open.
Second question: This is not correct. Assuming that $(x_n)_{n\in\mathbb N}$ is not a Cauchy sequence doesn't allow you to deduce that it satisfies the property mentioned in the statement. If it is not a Cauchy sequence, then there is a $\varepsilon>0$ such that$$(\forall p\in\mathbb{N})(\exists m,n\in\mathbb{N}):m,n\geqslant p\wedge d(x_m,x_n)\geqslant\varepsilon.$$It follows from this that there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ such that the distance between any two distinct elements of that subsequence is greater than or equal to $\varepsilon$. Now you can apply your argument.