Locally path connected but not path connected $[0,1]\cup[2,3]$ is locally path connected but not connected. 
How to rigorously prove it is locally path connected ?
 A: The starting point is that any interval in $\mathbb{R}$ (with the standard topology) is path connected (open, closed, half closed, bounded, unbounded, whatever).
Recall that a topological space $X$ is locally path connected if for any $x\in X$ and any open neighbourhood $V\subseteq X$ of $x$ there is a path connected open neighbourhood $U\subseteq X$ of $x$ such that $U\subseteq V$. Or equivalently: the collection of open path connected subsets is a basis for the topology.
Now let $X=[0,1]\cup [2,3]$ with the standard topology I assume. The topology on $X$, being derived from the standard topology on $\mathbb{R}$, is generated by "relative" open intervals. So for example $\big(\frac{1}{3}, \frac{2}{3}\big)$ is open and $\big[0,\frac{1}{2}\big)$ is open, but $\big[\frac{1}{3}, \frac{2}{3}\big)$ is not open.
Take any $x\in X$ and any open neighbourhood $V\subseteq X$ of $x$. There are two cases:


*

*$x$ is one of the endpoints $0,1,2,3$. Without loss of generality we can assume that $x=0$. Since $V$ is open then there is $\epsilon >0$ such that $[0,\epsilon)\subseteq V$ due to relative topology. Note that $[0,\epsilon)$ is an open neighbourhood of $x$ in $X$ and it is path connected because it is an interval. And it is fully contained in $V$.

*If $x$ is not one of the endpoints then there is $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subseteq V$ again due to relative topology. And as before $(x-\epsilon, x+\epsilon)$ is an open neighbourhood of $x$ fully contained in $V$ and path connected.
All in all $x$ has a path connected open sub-neighbourhood of $V$. By the arbitrary choice of $x$ and $V$ we conclude that $[0,1]\cup[2,3]$ is locally path connected.
Side note: there are other weaker variants of local path connectedness, for example that every point has an open path connected neighbourhood. And indeed, locally path connected does imply weakly locally path connected. But not the other way around as the comb space shows (or more generally: every path connected space is weakly locally path connected). Anyway usually those weaker versions are less useful (most theorems will assume the strong version implicitely).
A: Let $x \in [0,1] \cup [2,3]$, then ether $x \in [0,1]$ or $x \in [2,3]$.
Without loss of generality, assume $x \in [0,1]$ (the other case works in the same way). 
Let $U$ be an open neighborhood of $x$, it means that $U$ contains an open set which contains $x$. Recall that open sets in $\mathbb{R}$ always contain an open interval. Hence your neighborhood $U$ of $x$ will certainly contain an interval of the form $(x - \delta, x + \delta)$.
Therefore, in the subspace topology of $[1,2] \cup [2,3]$, a neighbourhood of $x$ will always contain a set of the form $(x - \epsilon, x+\epsilon)$, if $x$ is in the interior, or $[a,\epsilon)$ if for example $x=a$, i.e. on the boundary of your set. All these are path connected, and we are done.
Note: As pointed out in the comments, the notion of 'locally path connected' is different from others'locally something' property, in the sense that your property (path connectedness) has to be satisfied on every open neighborhood (i.e. every neighborhood has to contain a path connected one). It is not enough that there exists one satisfying your property. 
A: For every $x \in [0,1]$, $[0,1]$ is a locally path-connected (relatively) open neighbourhood of $x$. The same holds for $x \in [2,3]$ and $[2,3]$. 
Note that intervals in the reals are both connected and locally path-connected. 
