Invariance of Topological Manifold Dimension: $1$ Dimensional Case

I am currently practicing with a few problems in Lee's Introduction to Topological Manifold, and I decided the complete the following problem:

Prove that a nonempty topological space cannot be both a $1$-manifold and an $n$-manifold for $n>1$.

The idea of proof seems pretty obvious: Assume $X$ is both a $1$-manifold and an $n$-manifold for $n>1$. Given some $p \in X$, let $(U,\varphi)$ and $(V,\tau)$ be coordinate charts around $p$ where $\varphi:U \rightarrow \mathbb{R}$ and $\tau:V \rightarrow \mathbb{R}^n$. Note that both $U$ and $V$ are connected. Since $U$ and $V$ are both open, $W=U\cap V$ is an open, connected and nonempty subset of $X$. The restriction of $\varphi$ and $\tau$ to $W$ is a continuous map into $\mathbb{R}$ and $\mathbb{R}^n$, respectively. In particular, $\varphi(W)$ is a connected, open subspace of $\mathbb{R}$, hence $\varphi(W)=(a,b)$ for some $a,b \in \mathbb{R}$. As such, there exists some homeomorphism $f:(a,b)\rightarrow \mathbb{R}$. Using this we can construct the homeomorphism $f \circ \varphi|_W: W \rightarrow \mathbb{R}$. Similarly, we can construct a homeomorphism $g \circ \tau|_W: W \rightarrow \mathbb{R}^n$. Composing both of these maps in the right way gives a homeomorphism between $\mathbb{R}$ and $\mathbb{R}^n$, which is impossible since $\mathbb{R}$ minus a point is disconnected while $\mathbb{R}^n$ minus a point is connected for $n>1$.

My question is:

Is this line of reasoning correct?

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You could remark that there is no homeomorphism between $\mathbb{R}$ and $\mathbb{R}^n$ because $\mathbb{R}$ minus a point is disconnected, but $\mathbb{R}^n$ minus a point is connected. For other cases, comparing $n$ and $m$ dimensional manifolds, you need a generalization of this, Brouwers invariance of domain.
I suppose I left it out because the problem just before this one (which I also completed) was to show there was no homeomorphism between any open subset of $\mathbb{R}$ and $\mathbb{R}^n$ for any $n>1$. For the sake of completeness I will add it. – Holdsworth88 Aug 15 '12 at 10:11