Let H be a Hilbert space. Suppose X a subset of H.

Question

Let $H$ be a Hilbert space. Suppose $X$ a subset of $H$.

If $X$ is connected and compact then $X$ is connected by path?

Answer 1

Not true even for subsets of $\mathbb{R}^2$. If $G$ is a graph of $y=\sin 1/x$, $0<x\leq 1$, regarded as a subset of $\mathbb{R}^2$, then $X$ being a closure of the the graph $G$ is compact, connected, but not path connected. Is you draw a picture you will see it is not path connected. Proving it is connected is standard.

Answer 2

This fails in $\mathbb{R}^2$.

Let $X$ be the graph of $\sin(1/x)$ for $x\in (0,1]$, together with the segment $\{0\}\times[-1,1]$. Then $X$ is connected but not path connected.

For proofs see Theorem 3.7 and Corollary 3.8 of these notes:

https://kconrad.math.uconn.edu/blurbs/topology/connnotpathconn.pdf

Edit: Note that the set $X$ is precisely the closure of the graph of $y=\sin(1/x)$ over $(0,1]$. So this answer is the same as Piotr Hajlasz's (which came in just before mine, but I don't know if that means I should delete. )