Consider $X = \mathbb{R}P^2 \sqcup [0, 1]/ \sim$ where, for some pair of distinct points $x, y \in \mathbb{R}P^2$, we have the identifications $x \sim 0$ and $y \sim 1$.
The goal is to find the fundamental group of $X$ using Van Kampen.
So I was thinking of expressing it in terms of two open sets. Call the first one $A_1$. I was thinking of just letting this be $\mathbb{R}P^2$ itself (is that a problem?). Then let the other one, $A_2$ be $[0, 1]$ along with a path connecting $x$ and $y$ (i.e, $1$ and $0$). So the intersection would be the path between $x$ and $y$ which is certainly path connected, and (I believe) contractible; it's basically $[0 , 1]$ (I am dubious of this however). Then the only other unknown is $A_2$ which I want to say is just $S^1$, but I feel like that's wrong. In which case I would obtain $\pi_1(X) = \mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}$ which is probably incorrect (that's the impression I get from the follow up question).
I expect that I have misclassified $A_1 \cap A_2$ and $A_2$ so I would appreciate some help.
Edit: I just realized that $A_2$ isn't open. That's very dumb of me. Would it be fixed if I thicken the path into some open strip? This deformation retracts into a single path anyway so I'm not sure if it changes the story.
Your argument sounds good to me. You can pull back everything to $S^2$ to picture things better, then the situation looks like this:
As long as you thicken your $A_1$ to include some open neighborhoods of 0 and 1 (up to the blue lines in the picture), $A_1$ is open in the space, and as long as you thicken your $A_2$ to include an open (purple) neighborhood of the path between $x$ and $y$ in $S^2$, $A_2$ is open in the space. Also we have to pick a simple path from $x$ to $y$, e.g. a path passing through the hemisphere would probably yield complications down in $\Bbb{R}P^2$.