Theorem: If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof Attempt: Let $X$ be a topological space and $P$ be a path component of $X$. Note $\forall x_1,x_2 \in P$, there exist a continuous function $f:[a,b]\subset\Bbb{R}\to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Since $[a,b]$ is connected, then $f([a,b])$ is also a connected subspace of $X$ and $x_1,x_2 \in f([a,b])$. Thus, $P\subseteq f([a,b])$. Since connnected subspaces lie entirely within a component, let $f([a,b])\subset C$ where $C$ is some component of $X$. Hence, $P\subset C$.
Moreover, what if we also assume that $X$ is locally path-connected. So far we have $x_1,x_2 \in P$ and $f(a)=x_1, f(b)=x_2 \in f([a,b])$. Let $x'\in f([a,b])$. We want to show $f([a,b])\subset P$. Then, by local path-connectedness of $X$, for every open set $O'$ in $X$ such that $x' \in O'$ imply there exists a path-connected neighborhood $\mathscr{P'}$ such that $x'\in \mathscr{P'}\subset O'$. Since $x'\in f([a,b])$, then $\exists c\in [a,b]\subset\Bbb{R}(f(c)=x')$. Since $f$ is continuous, then $g=f|_{[a,c]}$ and $h=f|_{[c,b]}$ are continuous and by definition of a path, $g(a)=x_1$, $g(c)=x'$, $h(c)=x'$, and $h(b)=x_2$ imply there exist paths between $x_1$ and $x'$ and also $x'$ and $x_2$; denoted, $x_1\sim x'$ and $x'\sim x_2$ where $\sim$ relates two point linked by a path. Since $\forall\alpha\in\mathscr{P'}(\alpha\sim x')$, then, by transitivity of $\sim$, $\alpha\sim x_1$ and $\alpha\sim x_2$. By the symmetry of $\sim$, $\mathscr{P'}\subset P$, $P\subset\mathscr{P'}$, $f([a,b])\subset\mathscr{P'}$, and $\mathscr{P'}\subset f([a,b])$. Therefore, $f([a,b])=P$.
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 \cap U_2$ and so on, step by step, to b.
Thus K is path connected.