As requested by ronno, here is my comment as a full answer with more details (like both directions) and references (as a more general view on the concepts might be helpful). For completeness, I have also included the backwards direction:
Lemma: $X$ connected $\Leftarrow X$ path-connected
(The backwards direction does indeed not need the condition of $X$ being locally path-connected.)
Proof: Assume $X$ is path-connected, but not connected, then there are non-empty, disjoint and open subsets $U,V\subset X$ with $U\cup V=X$. Since they are non-empty, there are points $x\in U$ and $y\in V$ and since $X$ is path-connected, there is a path $\gamma\colon[0;1]\rightarrow X$ with $\gamma(0)=x$ and $\gamma(1)=y$. Consider $\gamma^{-1}(U),\gamma^{-1}(V)$:
- They are non-empty as $0\in\gamma^{-1}(x)\subseteq\gamma^{-1}(U)$ and $1\in\gamma^{-1}(y)\subseteq\gamma^{-1}(V)$.
- They are disjoint as $\gamma^{-1}(U)\cap\gamma^{-1}(V)=\gamma^{-1}(U\cap V)=\gamma^{-1}(\emptyset)=\emptyset$.
- They are open as preimages of open sets under a continuous map.
- They fulfill $\gamma^{-1}(U)\cup\gamma^{-1}(V)=\gamma^{-1}(U\cup V)=\gamma^{-1}(X)=[0;1]$
Hence $[0;1]$ would not be connected, which is not the case and hence yields a contradiction. $\square$
The last step has to be proven the hard way: If it could be parted into two non-empty disjoint and open subsets $U,V\subset[0;1]$ with $U\cup V=[0;1]$ (with $0\in U$ without loss of generality), then we would get the contradiction $\sup U\in U$ and $\sup U\in V$, so $U\cap V\neq\emptyset$.
Be aware: It might be tempting at first, but $[0;1]$ being connected must not be proven using that it is obviously path-connected and therefore connected because of the upper lemma as this is a circular conclusion.
Lemma: $X$ connected and locally path-connected $\Rightarrow X$ path-connected
Proof: Let $x\in X$ be a point and denote by $U=[x]_\sim\subseteq X$ its path-connected component, where $x\sim y$ iff there is a path $\gamma\colon[0;1]\rightarrow X$ with $\gamma(0)=x$ and $\gamma(1)=y$. (The relation is reflexive because of the constant path $\epsilon_x\colon[0;1]\rightarrow X,t\mapsto x$, symmetric because of the inverse path $\overline\gamma\colon[0;1]\rightarrow X,t\mapsto\gamma(1-t)$ and transitive because of the path composition.)
Let $y\in U$, so there is a path $\gamma$ from $x$ to $y$. Since $X$ is locally path-connected, there is an open and path-connected neighborhood $V$ of $y$, so for every point $z\in V$, there is a path $\delta$ from $y$ to $z$. With path composition, $\gamma*\delta$ is a path from $x$ to $z$, hence $V\subset U$ and $U$ is open.
Let $z\in\overline U$. Since $X$ is locally path-connected, there is an open and path-connected neighborhood $V$ of $z$ with $U\cap V\neq\emptyset$. Choose $y\in U\cap V$. Because of $y\in U$, there is a path $\gamma$ from $x$ to $y$ and because of $y\in V$, there is a path $\delta$ from $y$ to $z$. With path composition, $\gamma*\delta$ is a path from $x$ to $z$, hence $z\in U$ and $U$ is closed, meaning $U^\complement$ is open.
We get the partition $X=U\cup U^\complement$ into disjoint open subsets and since $X$ is connected, they cannot both be non-empty. Since $x\in U$ (as $\sim$ is reflexive) and $U$ is therefore non-empty, $U^\complement=\emptyset$ must be empty, so $X=U$, which means, that $X$ is path-connected. $\square$
Be aware: Since I sometimes encounter people believing the following myth, I want to add an important remark for dealing with local properties in topology: (Path-)connectedness does not imply local (path-)connectedness! The Warsaw sine curve is connected, but not locally connected (See "Counterexamples in Topology" by Steen and Seebach found here, Remark 6 on page 138.) The Comb space is path-connected, but not locally path-connected. (It is also contractible, but not locally contractible.)
Best Answer
Your proof doesn't work if $W_x$ is just some open set containg $x$ because such a set need even be connected.
Fix $x$ and consider the set of all points $y$ such that there is path in $V$ from $x$ to $y$. Since $V$ is connected it is enough to show that this set is both open and closed in $V$. [For the it is empty of=r equal to $X$. But the set contains $x$ so it cannot be empty].
If a take a point $y$ in this set then there is an open ball $B$ centered at $x$ contained in $V$. Let $z \in B$ There is a path from $x$ to $y$ and there is a path from $y$ to $z$ in $V$, namely $t \to tz+(1-t)y$. Hence there is a path in $V$ from $x$ to $z$. We have proved that the ball $B$ is contained in our set is it is open.
The fact that this set is close is proved in a similar way. (If there is no path from $x$ to $y$ there cannot be any path from $x$ to $x=z$ either!).
What is crucial to this proof is convexity of open balls in an normed linear space.