I think the answer is yes. See the example below: we consider the open interval $(-1,\infty)$ as an open submanifold of the smooth manifold $(i,\mathbb{R})$ with $i:\mathbb{R}\rightarrow \mathbb{R},i(x)=x$ being the chart. Similarly, $(i,\mathbb{R^2})$ is a smooth manifold with $i:\mathbb{R}^2\rightarrow\mathbb{R}^2,i(\textbf{x})=\textbf{x}$ being the chart.
Let $\varphi:(-1,\infty)\rightarrow\mathbb{R}^2$ by $\varphi(t)=\left(\frac{3t}{1+t^3},\frac{3t^2}{1+t^3} \right)$. The trace of $\varphi$ (with subspace topology) is the portion of the "folium of Descartes" lying in the first and second quadrants together with the origin. Observe that $\varphi$ is injective but near the origin, $\varphi^{-1}$ is not continuous (since $\varphi(0)=\textbf{0}$ but $\varphi(t)\rightarrow\textbf{0}$ as $t\rightarrow \infty$).
Hence, an injective smooth map between smooth manifolds can have a discontinuous inverse.
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You are correct. The preimage of any neighborhood of $0$ under $\phi^{-1}$, which is to say the image of $(-\epsilon,\epsilon)$ under $\phi$, will fail to be open in the subspace topology, since any neighborhood of $(0,0)$ in the subspace topology contains a point $\phi(t)$ with $t>\epsilon$.
Note that not only is your map smooth and injective, it is even an immersion.
This is a good example of how injective immersions can fail to be embeddings when their domain isn't compact.
Another classic example is $\Bbb{R}\to T^2 = \Bbb{R}^2/\Bbb{Z}^2$ given by $t\mapsto (t,\alpha t)$, where $\alpha$ is irrational. You can check that this is an injective immersion, but the open subsets of the image in the subspace topology have unbounded preimage, so the image of a bounded open subset of $\Bbb{R}$ is never open in the image.
See the image on wikipedia here to get a sense of this last example.