Let $p: \tilde{X} \rightarrow X$ be a covering map in the broad definition (empty sum allowed, as Hatcher does). First, $p[\tilde{X}]$ is open in $X$: let $y \in X$ with $x \in \tilde{X}$ such that $p(x) = y$. Let $U$ be an evenly covered neighbourhood of $y$, so $p^{-1}[U] = \sum_{i \in I} O_i$ (disjoint sum over index set $I$). As $x$ is in the sum, it is non-empty: some $O_{i_{0}}$ contains $x$ and $p |_{O_{i_0}}$ is a homeomorphism between $O_{i_{0}}$ and $U$. In particular $U \subset p[\tilde{X}]$, so $y$ is an interior point of $p[\tilde{X}]$.
Suppose $y \notin p[\tilde{X}]$. Then again take an evenly covered neighbourhood $U$ of $y$: $p^{-1}[U] = \sum_{i \in I} O_i$, disjoint sum over some index set $I$, where for every $i \in I$, the map $p|_{O_i}$ is a homeomorphism between $O_i$ and $U$. This means (!) that $I = \emptyset$, as otherwise we'd have a preimage for $y$, contradicting how we picked $y$. So in fact $p^{-1}[U] = \emptyset$, and this shows that $U \subset X\setminus p[\tilde{X}]$, so $p[\tilde{X}]$ is closed.
Now if $X$ is connected and $\tilde{X}$ is non-empty, then $p[\tilde{X}]$ is closed, open and non-empty, so equals $X$ by connectedness. So $p$ is surjective.
These are the only conditions we need: $X$ connected and $\tilde{X} \neq \emptyset$.
2,3)) Suppose to the contrary that a set $X=X_1\cup X_2\cup X_3$ is path-connected. Then there exists a continuous map $f(X)\to [0,1]$ such that $f(0)=(0,0)$ and $f(1)=(0,1)$. Put $X_-=\{(x,y)\in X:y\le 1/2\}$. Since the set $f^{-1}(X_-)$ contains $0$, it is non-empty. Since $X_-$ is a closed subset of $X$ and the map $f$ is continuous, a set $f^{-1}(X_-)$ is a closed subset of $[0,1]$, and so $f^{-1}(X_-)$ is compact. Thus a set $f^{-1}(X_-)$ contains its supremum $T$. Since $f(1)\not\in X_-$, $T<1$. It is easy to see that a set $X\setminus X_-$ splits into connected components which are the arcs of the graph and the set $X_3$. Since a set $(T,1]$ is connected, its continuous image $f((T,1])$ is connected too. Since $(0,1)\in f((T,1])\subset X\setminus X_- $, the only possibility to keep $f((T,1])$ connected is to have $ f((T,1]=\{(0,1)\}$. By the continuity of $f$, the set $f^{-1}(0,1)$ is closed, so it contains $T$. Thus $f(T)=(0,1)$, a contradiction with $f(T)\in X_-$.
3,4)) The set $X_1\cup X_2$ is path-connected, being a union of two intersecting path-connected sets (a segment and a graph of a continuous function on an interval).
Best Answer
The meaning of the phrase "sets $A, B$ meet" is that $A \cap B \ne \emptyset$.
We distinguish various cases.
Case 1: $X_1 = \emptyset$. Then the assertion is wrong because $\emptyset$ is a path component of $X_1$ which does not meet $X_2$. However, note that it is a matter of taste if $\emptyset$ is regarded as path-connected or not. Some authors do not, but the usual definition says it is (so that $\emptyset$ has a single path component, namely $\emptyset$ itself). I adopted this point of view, but if you don't like to do so the assertion becomes trivially true because then $\emptyset$ does not have any path component.
Case 2: $X_1 \ne \emptyset$.
Case 2.1: $X_2 = \emptyset$. Then the assertion is wrong.
Case 2.2: $X_2 \ne \emptyset$.
Case 2.2.1: $int X_2 = \emptyset$. Then the assertion is trivially true because this implies $X_1 = X$ (as pointed out in the question).
Case 2.2.2: $int X_2 \ne \emptyset$. This is the only non-trivial case.
To prove the assertion we have to show that for each $x \in X_1$ there exists a path $v : [0,a] \to X$ such that $v([0,a]) \subset X_1$, $v(0) = x$ and $v(a) \in X_2$.
Case a: $x \notin int X_1$. Then $x \in int X_2$ and we can take any constant path $v(t) \equiv x$.
Case b: $x \in int X_1$. Choose $y \in int X_2$. Since $X$ is path-connected, there exists a path $u : [0,b] \to X$ such that $u(0) = x, u(b) = y$. If $u([0,b]) \subset int X_1$, we are ready. Otherwise define $c = inf \lbrace t \in [0,b] \mid u(t) \notin int X_1 \rbrace$. We have $c > 0$ and $u([0,c)) \subset int X_1$. $u(c)$ cannot be contained in $int X_1$ (otherwise a neighborhood of $c$ would be mapped into $int X_1$ which would contradict the definition if $c$). Hence $u(c) \in int X_2$ and we conclude that a neighborhood of $c$ is mapped into $int X_2$. But then we find $a \in (0,c)$ such that $u(a) \in int X_2$. This means that $v = u \mid_{[0,a]}$ is a path as required.