Cantor set with pairs of points identified

Consider the middle-thirds Cantor set in $[0,1]$. I want to identify points in the following way.

First, identify the two points $1/3$ and $2/3$.

Then, idenfity $1/9$ with $2/9$, and also identify $7/9$ with $8/9$.

At the third step there will be four pairs of points: $1/27\sim 2/27$; $7/27\sim 8/27$; $19/27\sim 20/27$; $25/27\sim 26/27$.

Continue.

Essentially I am squeezing together the consecutive gaps in the Cantor set. I would like to know of the resulting space is homeomorphic to $[0,1]$.

Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=\sum_{n=1}^\infty\frac{2a_n}{3^n}\in K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=\sum_{n=1}^\infty\frac{a_n}{2^n}.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:K\to [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.