Does an uncountable discrete subspace of the reals exist? Does an uncountable and discrete subspace of the reals exist?
 A: Edit: The following "proof" is not correct, as pointed out in the comments. Carl Mummert and Asaf Karagila have given arguments which seem correct to me.
The answer in no.
Let $M\subset\mathbb R$ be an uncountable subspace. Then there is $n\in\mathbb N$ such that $M\cap [-n,n]$ is infinite (otherwise $M$ was countable).
Since $[-n,n]$ is compact, the infinite set $M\cap [-n,n]$ has an accumulation point in $[-n,n]$ and hence in $\mathbb R$. Therefore, $M$ cannot be discrete.
A: No. Take an uncountable subset of real numbers, then it has to have some $k$ such that there are uncountably many lie in $[k,k+1]$. 
Inductively we define smaller and smaller intervals in which there are uncountably many elements. These can be taken as closed, and their intersection will be nonempty by Cantor's theorem. 
Therefore every uncountable set of reals has an accumulation point and thus is not discrete. 
A: Several of the answers given rely on completeness or compactness, but these properties are not needed. For example there is no uncountable discrete subspace of the irrationals. The reals are a second-countable space, so any subspace is also second-countable, which prevents the subspace from having an uncountable discrete subspace. 
The proof is as follows. Suppose that $A$ is a discrete space with a countable basis $(B_j : j \in \mathbb{N})$. We want to prove $A$ is countable. Discreteness of $A$ means that each $a \in A$ is in some basic open set that contains $a$ and no other point in $A$. Thus we can make a map $f$ which sends each $a \in A$ to the least $j \in \mathbb{N}$ such that $B_j = \{a\}$. This is an injection from $A$ to $\mathbb{N}$, so $A$ is countable. 
A: Every uncountable set of the reals contains a limit point of itself. In fact, if $A$ is uncountable, all but countably many points of $A$ are limit points of $A$.
See, e.g.,  here
Of course, discrete sets do not contain any of their limit points.
A: Assume $X\subseteq\mathbb{R}$ is uncountable and discrete, ie. every point $x\in X$ is isolated. Thus for each point $x\in X$ there is an open neighbourhood $U_x$ such that $U_x\cap X=\{x\}$. Now, using this and the ordering of $\mathbb{R}$ we can find cover of $X$ consisting of open disjoint sets $\{U_x\}_{x\in X}$ st. $U_x\cap X=\{x\}$ for each $x\in X$. Since $\mathbb{R}$ is separable, this collection must be countable, and hence $X$ is countable, which contradicts the assumption.
A: Let $S$ be a discrete Subset of $\mathbb R$. By definition $S$ is closed. Since $\mathbb R$ is $\sigma$-compact, that is $\mathbb R = \bigcup_{i=1}^{\infty} K_i$ with $K_i$ compact subspaces of $\mathbb R$, we have that $S \cap K_i$ is compact and discrete, hence finite, so $S = \bigcup_{i=1}^{\infty} S \cap K_i$ must be countable.
A: Suppose that $A\subseteq \mathbb{R}$ is discrete.  Then, for each $a\in A$, there exists an open interval $I_a\subseteq \mathbb{R}$ such that $I_a\cap A=\{a\}$.  Since $\mathbb{Q}$ is dense in $\mathbb{R}$, pick (via the Axiom of Choice, if necessary) a rational number $q_a$ in each $I_a$ such that $q_a\notin I_b$ for each $b\in A\setminus\{a\}$.  Then, the mapping $f:A\rightarrow \mathbb{Q}$ defined by $f(a)=q_a$ is injective, implying that $A$ is countable.
