Simple proof that downward intersections of simply connected compact sets are simply connected

Question

Let $X$ be a topological space, and $S_0, S_1, \dotsc \subset X$ be simply connected compact sets with $S_{n+1} \subset S_n$.

Question: Is there a simple proof that $S = \bigcap_n S_n$ is simply connected, for nice enough $X$?

If $X$ is the plane, at least in the case I care about the result follows from Alexander duality. But this seems annoyingly heavyweight.

If $C([0,1]^2, X)$ were compact the result would follow immediately by choosing a limit point of a sequence of homotopies, but it isn’t compact. Is it possible to rescue this kind of simple proof, if $X$ is nice?

Some intuition for why I think this might be possible: if we have an open ray $r : [0,1) \to X$ for compact $X$, the set of limit points of $r$ at 1 is connected even if there are many limit points. Similarly, if we pick a sequence of homotopies, a subsequence converges pointwise to a possibly discontinuous map, but each tear introduced should somehow be connected in $S$.

Edit: As discussed in the comments, the statement is false as exhibited by the topologist's sine curve. I was confusing "simply connected" with "full", and indeed the result for "full" follows from unions of jointly intersecting connected sets being connected.

Answer 1

Several commenters observed that the topologist's sine curve is a non-path connected subspace of $\mathbb{R}^2$ that can be written as an intersection of countably many simply connected spaces. One can modify this construction to obtain an example of a path connected space that is not simply connected but which is the intersection of countably many simply connected spaces.

We observe however that the intersection of countably many connected compact Hausdorff spaces is also connected compact and Hausdorff. Therefore, the reason that the notion of simple connectedness is not closed under countable intersection is that simple connectedness is too reliant on the notion of a path and paths may not always squeeze through countable intersections of closed sets, but if we can generalize the notion of simple connectedness in such a way that we no longer have so much reliance on paths, then we may obtain a variation of simple connectedness that is closed under countable intersection.

We say that a compact Hausdorff space $X$ is intersectionally simply connected if it is connected and there is a larger topological space $Z$ and a decreasing collection of simply connected spaces $Z_n\subseteq Z$ with $X=\bigcap_{n\in\omega}Z_n$.

Proposition: The intersection of countably many intersectionally simply connected spaces is also intersectionally simply connected.

Proof outline: Suppose that $(X_n)_{n}$ is a decreasing sequence of intersectionally simply connected spaces with $X=\bigcap_nX_n$. Since the intersection of compact Hausdorff connected spaces is compact Hausdorff and connected, we know that $X$ is compact Hausdorff and connected. For each $n\in\omega$, there is some larger space $Z_n$ and a decreasing sequence of simply connected subspaces $Z_{n,m}\subseteq Z_n$ such that $X_n=\bigcap_{m\in\omega}Z_{n,m}$.

Now, consider the disjoint union $Y=\bigcup_{n\in\omega}(Z_n\times\{n\})$, and the quotient space $Y/{\simeq}$ where we set $(x,i)\simeq(x,j)$ whenever $x\in X$ but where we have no other relations other than equality. Then let $Y_m=\bigcup_{n\in\omega}(Z_{n,m}\times\{n\})$. Then $(Y_m/{\simeq})\subseteq (Y/{\simeq})$ and each $Y_m$ is simply connected, but $\bigcap_{m\in\omega}(Y_m/{\simeq})$ is isomorphic to $X$. Therefore, $X$ is intersectionally simply connected. $\square$