I am trying to show that the locus of points $(x,y) \in \mathbb{R}^2$ such that $xy = 0$ is path connected. That is, given points $(a,b), (c,d)$ in this set, I must define directly a function $f: [0,1] \to \mathbb{R}^2$ that is continuous and for which $f(0) = (a,b)$ and $f(1) = (c,d)$ and for which $f(t) \in \{(x,y) \mid xy = 0\}$ for all $t \in [0,1]$.
My problem is with formalizing the argument. The set is just the union of the $x$ and $y$ axes. Surely $(0,0)$ is an element of this set. Given anything on the $y$-axis, I plot a vertical segment from that point to $(0,0)$, either from above or below. Given anything on the $x$-axis, I plot a horizontal segment from that point to $(0,0)$. By composing paths, because I've already shown path connectedness gives an equivalence relation, I can join any two points in the set.
The problem is formalizing this by defining the function explicitly. It should suffice to show that given $(a,b)$ in the set (so one component is zero), I have $(a,b) \sim (0,0)$: by symmetry, transitivity, the result should immediately follow. I need to consider cases $a > 0$, $a < 0$, $b > 0$, $c < 0$. I can dr aw the segment, but I don't know how to parametrize it directly.
Any ideas would be appreciated.
Let $A$ be your set.
If $x\in\Bbb R$, define $\gamma_x^h(t)=(tx,0)$, for each $t\in[0,1]$. That's a path in $A$ going from $(0,0)$ to $(x,0)$.
Now, if $y\in\Bbb R$, define $\gamma_y^v(t)=(0,ty)$. That's a path in $A$ going from $(0,0)$ to $(0,y)$.
So, I have proved that, for each $p,q\in A$, there is a path $\alpha$ in $A$ from $(0,0)$ to $p$ and there is a path $\beta$ in $A$ from $(0,0)$ to $q$. The domains of both pathts are $[0,1]$. So, define$$\begin{array}{rccc}\gamma\colon&[0,2]&\longrightarrow&A\\&t&\mapsto&\begin{cases}\alpha(1-t)&\text{ if }t\in[0,1]\\\beta(t-1)&\text{ if }t\in[1,2],\end{cases}\end{array}$$and $\gamma$ will be a path in $A$ going from $p$ to $q$.