Why a connected subspace of a locally connected space X is locally connected if X is the real line? Why a connected subspace of a locally connected space X is locally connected if X is the real line?
Is this true if X is an arbitrary locally connected space?
Thanks for your help
 A: This is true if $X=\mathbb{R}$ because connected subsets in this case are precisely the intevals.
For a counterexample for other spaces take $X=\mathbb{R}^2$ and $Y=\{ (x,y): x=0,\ 0\leq y\leq 1\} \cup \{ (x,y) :x\in [0,1],\ y=\sin\left(\frac{1}{x}\right) \}=Y_1\cup Y_2$. Then points in $Y_1$ don't have connected open neighbourhoods (in $Y$).  
A: HINTS:


*

*Every non-empty connected subset of $\Bbb R$ is homeomorphic to one of four subsets of $\Bbb R$. Two of these are $\{0\}$ and $\Bbb R$ itself. The other two are also very simple and familiar; what are they? Once you’ve found them, you should be able to see easily that all four are locally connected, and therefore every subspace of $\Bbb R$ is locally connected.

*Let $X=\Bbb R^2$ and $S=\{0\}\cup\{1/n:n\in\Bbb Z^+\}$. Let $$Y=\Big([0,1]\times\{0\}\Big)\cup\Big(S\times[0,1]\Big)\;.$$ Show that $Y$ is connected. Is $Y$ locally connected? (Consider neighborhoods of the point $\langle 0,1\rangle$.)
The space in (2) is sometimes known as the comb space; it’s a useful source of examples involving connectedness-type properties.
