Let $X$ be the space and fix $p \in X$. Let $C$ be the set of all points in $X$ which are path connected to $p$. Now, it is enough to show that $C$ is closed and open in $X$, to show that $C=X$ (given that $p \in C$ via a constant path, so $C$ is non-empty).
To show $C$ is open, let $c \in C$, then we can choose (by local path connectedness) an open subset $U$ containing $c$. For $u \in U$, $u$ is path connected to $c$ which is path connected to $p$, so by joining paths, we have that $u$ is path connected to $p$. In other words, $U \subset C$, so $C$ is open.
Look at the closure of $C$, namely $\bar{C}$. Let $d \in \bar{C}$, and choose, like above, an open path connected subset $V$ containing $d$. Note that $V \cap C \neq \phi$, because $V$ is open! Hence let $e \in V \cap C$, then $d$ is path connected to $e$ which is path connected to $p$ (because $e \in C$)! Hence $d$ is path connected to $p$ and $d \in C$, so $C=\bar{C}$, $C$ is closed.
Hence $C=X$, and $X$ is path connected.
EDIT : In the style of equivalence relations , we can create another proof. Indeed, define an equivalence relation on $X$ given by $p \sim q$ if there is a path connecting $p$ and $q$. It is easy to see that this is an equivalence relation : for reflexivity, use the constant path. For symmetry, use the reverse of a path, and for transitivity, use the concatenation of the two paths.
Finally, note that every equivalence class is open, because if I take $p$, then by local path connectedness, some neighbourhood of $p$ is path connected, but this includes $p$. Therefore, this entire neighbourhood is in the same equivalence class as $p$. The openness follows.
Now, the equivalence classes partition $X$ into disjoint open sets. Since $X$ is connected, this can't happen unless there's only one equivalence class, which is the whole of $X$. We are done!
Best Answer
A simply connected set of course (by definition) is path-connected, but on the other side a path-connected and locally path-connected set need not to be simply connected. The easiest example is an open disk $B \subset \mathbb{R}$ around $0$, with an additional point $0'$, and adding to the topology base a base around $0$, with $0$ substituded with $0'$. Figuratively, $0$ is "doubled".
Any continuous homotopy to this space, starting with a path through $0$, cannot contain $0'$, because of contonuity (the preimage of the paths passing through $0'$ and of the paths passing through $0$ would disconnect $[0,1]$, which is bviously connected) so paths through $0$ and through $0'$ are not homotopically equivalent.