Suppose we have two distinct fixed points $x, y \in X$ where $X$ is a Banach space. If we define $\gamma: [0,1] \to X$ by $t \mapsto (1-t)x + ty$, I think the map $\gamma: [0,1] \mapsto \gamma([0,1])$ when restricted on the image is a homeomorphism.
The way I would to prove is to note the map is bijective and open since $t \mapsto t(x-y)$ is open by open mapping theorem and translation is a homeomorphism. Is this argument correct? Is there some more direct way to see this fact?
That works (assuming you are applying the open mapping theorem to the span of $x-y$) but the open mapping theorem is massive overkill. You can just observe that $\gamma$ is a closed map since $[0,1]$ is compact and $X$ is Hausdorff (if $A\subseteq[0,1]$ is closed then it is compact so $\gamma(A)$ is compact so $\gamma(A)$ is closed).
Or, even more simply, observe that the norm on the span of $x-y$ is just $\|t(x-y)\|=|t|\|x-y\|$. So $\gamma$ just multiplies distances by $\|x-y\|$. It follows that the inverse of $\gamma$ just multiplies all distances by $\frac{1}{\|x-y\|}$ and so is continuous.