Any set containing an unbounded interval is path-connected. The intuitiv idea is that you can walk to infinity and jump from there to any point you like.
E.g. if it contains $(y_0,\infty)$ for some $y_0\in \mathbb{R}_{>0}$. Let $x_0\in \mathbb{R}$. If $x_0\geq y_0$, then they are clearly connected by a continuous path ($\gamma:[0,1]\rightarrow X, \gamma(t)= x_0+ t(y_0-x_0$)). On the other hand, if $x_0< y_0$, then we have the path
$$ \gamma: [0,1] \rightarrow X, \gamma(t)=\begin{cases} \frac{y_0}{t},& t\neq 0, \\ x_0,& t=0. \end{cases} $$
Let me show that is continuous in the case $x_0\geq 0$ (the case $x_0<0$ is similar). Note that any open nbhd $U$ of $x_0$ in $X$ can be written as
$$ U= V \setminus \{x_0\} \cup (-\infty, -m) \cup (\{x_0\} \cup (n, \infty)) $$
where $m,n\in \mathbb{R}_{>0}$ and $V\subseteq \mathbb{R}$ open and bounded. Then we have
$$ \gamma^{-1}(U) = \gamma^{-1}(V\setminus \{x_0\}) \cup \gamma^{-1}((-\infty, -m)) \cup \gamma^{-1}(\{x_0\} \cup (n,\infty)).$$
We show that all of those sets are open. First we note $\gamma^{-1}((-\infty, -m))=\emptyset$, which is open in $[0,1]$. Next we have
$$ \gamma^{-1}(\{x_0\} \cup (n,\infty)) = \begin{cases} [0, \frac{y_0}{n}),& n>y_0, \\ [0,1],& n\leq y_0. \end{cases} $$
In both cases the sets are open in $[0,1]$.
Finally, as $V$ is bounded, there exists $R>0$ such that $V\subseteq (-R,R)$. Then we define
$$ \tau : [0,1] \rightarrow \mathbb{R},\tau(t):= \min \{ R, \gamma(t) \}. $$
As $\tau$ is continuous and $\tau^{-1}(V\setminus \{x_0\} ) = \gamma^{-1}(V\setminus \{x_0\})$, we get that also $\gamma^{-1}(V\setminus \{x_0\})$ is open in $[0,1]$.
Right now I do not have time to work it out, but I guess that that if a set does not contain an unbounded interval, then it is path-connected in $X$ iff it is path-connected in $\mathbb{R}$. The idea is that the space is first-countable and thus continuity and sequential continuity coinced (see here Sequentially continuous implies continuous). Then we should be able to use that bounded sequences in $X$ converge iff they converge in $\mathbb{R}$. This means you cannot jump unless you are at infinity.
I'm submitting a finalized version of my original proof as a proposed answer.
1. Theorem
The infinite product of connected spaces is connected in the product topology. In particular, let $ \{X_\alpha \}_{\alpha \in A}$ be an infinite collection of connected topological spaces. Then
$$ X := \prod_{\alpha \in A} X_\alpha $$
is connected, where $ (X, \tau_X) $ has the product topology.
2. Lemma
Fix $ x \in X $ and define $X_K \subseteq X$ for finite index set $K$ as
$$ X_K := \{ z \in X : \forall i \notin K, z_i = x_i \} $$
Then $x \in X_K$ and $ X_K $ is homeomorphic to $ \prod_{k \in K} X_k $.
3. Proof of Lemma
Define $ f : X_K \rightarrow \prod X_k $ as simply removing coordinates not in $K$.
This is clearly surjective. And injective: if $f(u) = 0$, then $u_k = 0$ for each $ k \in K $ (and the coordinates outside of $K$ are fixed).
Next, note that for any open $U \subset X_K$, as subspace of $X$, there exists in $V \in \tau_X$ such that $ U = V \cap X_K $. Thus, each $U_k$ is open in $X_k$ and the remaining coordinates are fixed.
Since any open $ W \subset \prod X_k $ has open $W_k$ in $X_k$, it's immediate that $f$ and $f^{-1}$ map $ U_k \leftrightarrow U_k $. It follows that f is a homeomorphism. $\square$
4. Proof of Theorem
Again fix $ x \in X $. Let $ E := C(x) $, the connected component of $x$ in $X$.
Since E is closed (we will not prove this here), it suffices to show that E is dense.
First note that $\prod X_K$ is connected for any for any finite $K$, since the finite product of connected sets is connected (not proven here). Now we use the Lemma to conclude that $X_K = f^{-1}(\prod X_k)$ is connected (connectedness is a topological property). It follows that $X_K \subseteq E $ for all finite $K$.
It remains to show that $ F:= \bigcup \{ X_K : finite~K \} $ is dense in X. This will follow from a property of the product topology. Also, note that the arbitrary union of connected sets is connected (not shown here), so $F \subseteq E$.
We will show that all points of $X$ are adherent to $X_K$ for some finite $K$ and then be done. Fix $y \in X$ and a y-neighborhood $U \in \tau_X$. From the product topology, we have
$ U = \prod U_\alpha $ with $U_\alpha \ne X_\alpha $ only finitely often with $ U_\alpha$ open in $X_\alpha$.
Then let $I$ be the finite set of indices where $U_\alpha \ne X_\alpha$.
Then we can choose a $z \in X_K$ with $K = I$ and $ z_k = k_k $ for each $ k \in K$.
It follows that $z \in U$ and all points of $X$ are adherent to $F \subseteq E$. $\square$
Best Answer
Your proof for the product topology is fine: you use the important and basic characterisation of continuous maps that map into a product, and this characterisation of continuity actually uniquely determines the product topology, as I wrote about here, e.g. As path-connectedness is defined by the existence of certain continuous maps, the proof of path-connectedness is quite natural.
For a finer topology nothing sensible can really be said, and the first natural finer candidate, the box-topology, fails miserably: no non-trivial box-product is connected, let alone path-connected. Also other properties fail for finer topologies, compactness being a particularly important one. That's why the product topology is usually the only suitable candidate and its properties guarantee the path-connectedness of a product of path-connected spaces.