I am looking for a yes or no answer to the following question, though if there is a simple explanation I'd like that as well:
If $X$ is any continuous image (not necessarily metrizable) of $[0,1]$, then is $X$ necessarily locally connected?
Just need a clarification here because a lot of times metrizability is implicitly assumed in the classical theorems.
Here's a simple counterexample. Let $X$ be $[0,1]$ with the topology that a proper subset $U\subset[0,1]$ is open iff it is a union of intervals of the form $(1/(n+1),1/n)$ for $n$ a positive integer, with the additional constraint that if $(1/2,1)\subseteq U$ then $(1/(n+1),1/n))\subseteq U$ for all but finitely many $n$. Clearly $X$ is a continuous image of $[0,1]$, since the identity map $[0,1]\to X$ is continuous. But $X$ is not locally connected at any point of $(1/2,1)$, since any neighborhood $U\neq X$ of a point of $(1/2,1)$ contains $(1/(n+1),1/n)$ for some $n>1$ and then $(1/(n+1),1/n)$ and $U\setminus (1/(n+1),1/n)$ is a nontrivial partition of $U$ into open sets.
Note however that if you require that $X$ is not just a continuous image but a quotient of $[0,1]$ (which is automatic if $X$ is metrizable), then $X$ must be locally connected. Indeed, suppose $q:[0,1]\to X$ is a quotient map, $x\in X$, and $U\subseteq X$ is a neighborhood of $x$. For each $t\in q^{-1}(\{x\})$, pick an open interval around $t$ contained in $q^{-1}(U)$, and let $V_0$ be the union of these open intervals. Note that $q(V_0)$ is connected, as a union of connected sets which all overlap at $x$. Now for each $t\in q^{-1}(q(V_0))$, pick an open interval around $t$ contained in $q^{-1}(U)$, and let $V_1$ be the union of these open intervals. Again, $q(V_1)$ is connected since it is a union of connected sets all of which intersect the connected set $q(V_0)$. Repeating this inductively, we get a nested sequence of open sets $V_n\subseteq q^{-1}(U)$ such that $q(V_n)$ is connected for each $n$ and $V_n$ contains all of $q^{-1}(q(V_{n-1}))$. Let $V=\bigcup V_n$. Then $V=q^{-1}(q(V))$, so $q(V)$ is open. Moreover, $q(V)$ is connected, contains $x$, and is contained in $U$. This shows that $X$ is locally connected.
(In fact, this argument more generally shows that any quotient of a locally connected space is locally connected.)
