# Help with a proof; Path connected open subspaces imply $X$ to be path connected

This is a problem I cannot prove,

Show that if $$X$$ is connected and is such that for every $$x \in X$$ there exists and open path-connected set $$U$$ with $$x \in U$$ then it is path-connected.

Here's my poor attempt

$$X$$ is connecetd so let some continuous function $$f:X \to D$$ to a discrete space exists. Then, this map is constant. Since $$U$$ is path-connecetd, it is also connected. Therefore, any restrcition $$f_U=f$$. i.e. The map is constant too.

Let $$x_1,x_2 \in U$$. Since it is path-connected, there exists some path $$\alpha:I \to U$$. Then, $$f(x_1)=f(x_2)=f(\alpha(0))=f(\alpha(1)) \in D$$. Since $$f$$ is continuous, and noting that a point in the discrete space is open by the discrete topology, $$f^{-1}\{f(\alpha(t)\} \in X$$ is open. In fact, this is the entire set $$X$$.

Now, I'm thinking that having elements/points from path-connecetd $$U$$ now related to the entire set $$X$$ has something to do with...showing $$X$$ is also path-connected. Or something. But I can't rigorously proceed from here.

Or maybe my entire thought process is wrong. I don't know what to do with the vast amount of information, connectedness, open subsets, path-connected subset, continuous map to a discrete space is constant etc etc.

Which ones I can use, which ones I don't need is just to complicated for me to guess. Please help

HINT: For $x\in X$ let $C(x)$ be the path component of $X$.
• Show that $C(x)$ is open for each $x\in X$. Here’s where you’ll use the hypothesis that each point has a path-connected open nbhd.
• Show that $\{C(x):x\in X\}$ is a partition of $X$.
• Use the connectedness of $X$ to conclude that $X=C(x)$ for each $x\in X$.