What I want to show for this is that $V \setminus \{x\}$ is homeomorphic to some punctured ball, $B_m \setminus \{p\}$ (where $B_m$, $p \in \Bbb R^m$).
And then since $B_m \setminus \{p\}$ is path-connected, $V \setminus \{x\}$ is path-connected hence connected.
So far all I've got is this: I know that since $X$ is an $m$-manifold there a homeomorphism $$\phi_\alpha : U_x \rightarrow A$$ where $U_x$ is a open neighborhood of x, $U_x \subseteq X$, and $A$ is open subset of $\Bbb R^m$.
Then take $C = \phi_\alpha(U_x)\cap A$. Then since $\phi_\alpha(U_x)$ is open and $A$ is open, $C$ is open. So then I can take a ball $B_m \subseteq C$. But this is where I get stuck.
You're very nearly there:
Since $A = \phi(U_x)$ is open, there is some ball $B \subseteq A$ containing $p := \phi(x)$ (because open balls form a basis of the standard topology on $\mathbb{R}^n$). If we denote $V := \phi^{-1}(B)$, the restriction $\phi\vert_V$ is a homeomorphism between $V \ni x$ and the ball $B$, and restricting further gives a homeomorphism between $\phi^{-1}(B - \{p\}) = V - \{x\}$ and the punctured ball $B - \{p\}$.