Proof:

Let $x$ and $y$ be arbitrary points of $X$ and let $G$ be the set of all points of $X$ that can be joined to $x$ by a path. Clearly $G$ is non-empty since it contains the point $x$. If we show that $G$ is open and closed then by connectedness of $X$ it would follow that $G$ would equal the whole space $X$. In particular $G$ contains $y$ thereby proving that there is a path in $X$ joining $x$ and $y$. First we show that $G$ is open. Well, let $z$ be an arbitrary point of $G$ and choose a path $\gamma :[0, 1] \longrightarrow X$ such that $\gamma(0) = x$ and $\gamma(1) = z$. Choose a path connected neighborhood $N$ of $z$ and $w \in N$ be arbitrary. Then there is a path $\sigma$ lying in $N$ joining $z$ and $w$. We now juxtapose $\gamma$ and $\sigma$ by defining $\eta :[0, 1]\longrightarrow X$ as

$$\eta(t) = \left\{\begin{array}{lll} \gamma(2t), & & 0 \leq t \leq 1/2 \\ \sigma(2t - 1), & & 1/2 \leq t \leq 1 \\ \end{array} \right.$$

By virtue of the gluing lemma $\eta$ is continuous and defines a path joining $x$ and $w$. Hence $w$ belongs to $G$ and so $N \subset G$. We now show that $G$ is closed as well. Let $y \notin G$ and $N$ be a path connected neighborhood of $y$. Then we show that $N\subset X-G$. Well, if not, pick $z \in G\cap N$ and there is a path $\gamma$ in $G$ joining $x$ and $z$ and a path $\sigma$ in $N$ joining $z$ and $y$. Juxtaposing we would get a path in $X$ joining $x$ and $y$ which would contradict the fact that $y \notin G$. So $X-G$ is also open in $X$ and the proof is complete.