The only topology similar to the Euclidean topology on $\mathbb{R}$ is the Euclidean topology.

Suppose there is such a topology $\tau$. I'll use "open," "continuous," etc. to mean with respect to the Euclidean topology and "$\tau$-open" etc. for $\tau$.

Since $\tau$ is a refinement of the Euclidean topology, there exists a point (without loss of generality $0$) and a $\tau$-open neighborhood $U$ which does not contain any open interval containing $0$. Then there is a sequence $x_i$ in the complement of $U$ converging to $0$. For convenience choose a monotonic subsequence $y_i$, without loss of generality positive, and suppose $y_1=1$.

Now consider the function $\mathbb{R}\to\mathbb{R}$ which fixes the complement of $(0,1)$, takes $[\frac{1}{2n},\frac{1}{2n-1}]$ to $y_n$, and takes $[\frac{1}{2n+1},\frac{1}{2n}]$ to $[y_{n+1},y_n]$ linearly. This function is continuous, hence $\tau$-continuous.

Then the preimage of $(-1,1)\cap U$ is a $\tau$-open subset $V$ of $(-1,0] \cup \bigcup_n (\frac{1}{2n+1},\frac{1}{2n})$ which contains $0$.

There is a homeomorphism (hence a $\tau$-homeomorphism) $\phi$ of $\mathbb{R}$ which fixes the complement of $(0,2)$, takes $[1,2]$ to $[\frac{1}{2},2]$ linearly, and takes $[\frac{1}{k+1},\frac{1}{k}]$ to $[\frac{1}{k+2},\frac{1}{k+1}]$ linearly.

Then $\phi(V)\cap V$ is a $\tau$-open subset of $(-1,0]$ containing $0$.

But then $(-\infty,0]$ is $\tau$-open, so that $\mathbb{R}$ is not $\tau$-connected, a contradiction since then there is a $\tau$-continuous map sending $(-\infty,0]$ to one point and $(0,\infty)$ to another.