The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a nice fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a nice map from the circle to $X$ representing the supposed generator of $\pi_1(X)$. (Namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$.) Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.

An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.

[Edited for clarity and to take into account editing of the original post.]

The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a map from the circle to $X$ representing the supposed generator of $\pi_1(X)$, namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$. Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.

An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.

The fundamental group of this space $X$ can be computed by applying the standard van Kampen theorem not to $X$ itself but to the mapping cylinder of a nice map from the circle to $X$ representing the supposed generator of $\pi_1(X)$. Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ regarded as a subspace of the mapping cylinder. Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.

An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.

The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a nice fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a nice map from the circle to $X$ representing the supposed generator of $\pi_1(X)$. (Namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$.) Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.

An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.

[Edited for clarity and to take into account editing of the original post.]