I am trying to showed that if $X$ and $Y$ are path connected then $X \coprod_f Y$ is path connected. (The adjunction space). Let $A \subset X$ , and let $f:A \to Y$ be the attaching map.
(Note: there is some abuse of notation in what follows, in terms of re-use of symbols for function names for different paths)
So path connectedness of $X,Y$ means for every $a,b \in X$ (or $Y$) there exists a continuous map $\phi:I \to X: \phi(0)=a, \phi(1)=b$ (or equivalent for $Y$)
Let $p: X \coprod Y \to X \coprod_f Y$ be the quotient map. Consider any two points $a,b$. If $p^{-1}(a)\cap Y = \emptyset,p^{-1}(b)\cap Y = \emptyset$ then path connectedness is clear (again, also replace $Y$ with $X$), as $X$ and $Y$ are path connected.
If $p^{-1}(a) \in A$ and $p^{-1}(b) \in A$ (I know that doesn't strictly make sense, but I think it is clear what is meant!) then path connectedness is again clear, as we can take the path given by $\phi: I \to A:\phi(0)=p^{-1}(a),\phi(1)=p^{-1}(b)$ and compose with $p$ (I guess via the maps $A \to X \to X \coprod Y \to X \coprod_f Y$) to give a path $\phi' :I \to X \coprod_f Y: \phi'(0)=a, \phi'(1)=b$
So say $p^{-1}(a)$ is in $X$, but not in $A$ and $p^{-1}(b)$ is in $A$. Then there is still a path $\phi:I \to X: \phi(0)=a, \phi(1)=b$. We can then break this into two paths $\phi_1:I \to X: \phi(0)=a, \phi(1)=t$, $\phi_2:I \to X: \phi(0)=t, \phi(1)=b$, where $(a,t) \in X$, $(t,b) \in A$, and then we can just form the path that is the composition of the two paths above (probably using the gluing lemma or something similar)
I think this argument could be made rigorous - but I also feel like there is a nice simple proof of this that I am missing.