Would a connected space contain a compact subspace

Question

I am trying to prove that in a connected space - $X$ , for every two elements of $X$ - say $a,b$ I can find a subspace of $X$ ( say $X'$ ) , such that$ X'$ contains a,b and is also connected, and compact.
I managed to do so if $X$ is metric or Hausdorff, or path connected. But for a general connected space I can't figure why it has to be true. Thanks guys!

Answer 1

Let $X=\{(x,y)\in\mathbb R\times\mathbb R:x\ne0\text{ and }y=\sin\frac1x\}\cup\{(0,0)\} $ with its induced topology as a subspace of $\mathbb R\times\mathbb R$. Then $X$ is a connected (but not pathwise connected) metric space. There is no connected compact subspace of $X$ containing $\{(0,0)\}$ and any other point. If $x_1\lt0\lt x_2$, there is no connected compact subspace of $X$ containing $(x_1,\sin\frac1{x_1})$ and $(x_2,\sin\frac1{x_2})$.

This example is a variant of the Topologist's Sine Curve. The official version of the Topologist's Sine Curve, namely $\{(x,y)\in X:0\le x\le1\}$, works just as well as a counterexample here.

Answer 2

If the space is path connected, you may take the smaller connected, compact space to be a path between the two points.