So, in section 1.3 of Hatcher, there is a particular exercise stated as follows:
9) Show that if a path-connected, locally path-connected space $X$ has $\pi_1(X)$ finite, then every map $X \to S^1$ is null-homotopic.
I have already gone through the work of proving this, but then I wondered the following: What would happen if we let $X = I$? I know that the unit interval is simply-connected, but if it were locally path-connected, then the above exercise would imply that every path in $S^1$ (and in particular loops) would be null-homotopic, contradicting the fact that $\pi_1(S^1) \cong \mathbb{Z}$. The only way I can see avoiding this is if $I$ were in fact not locally path-connected, but I can't see why.
Best Answer
$[0,1]$ is most certainly locally path connected - given any point $a$ and any open (or half-open for the endpoints) interval containing it is path-connected, since any two points $x,y$ are connected by a path $[x,y]$.
Your mistake lies in the fact that in the definition of $\pi_1(S^1)$ admits arbitrary homotopies, while actually it only admits homotopies which fix the endpoints. If you allow the endpoints of a loop to move (thus making it not a loop at some points in the homotopy), you can contract it to a point.