Let $S$ be a connected surface. Pick $x\in S$, then by definition there exists an open neighborhood $U_x\ni x$ of $S$ homeomorphic to an open disc $D(0,1)\subset \mathbb{R}^2$.
To exclude the case $S=\overline{U_x}$ mentioned in the comment, we consider the homeomorphic image $Z_x$ of $\overline{D(0,1/2)}$, which must be compact in the Hausdorff space $S$ so closed in $S$.
Moreover $D(0,1/2)$ is homeromorphic to the interior of $Z_x$ in $U_x$ which is the same as the interior of $Z_x$ in $S$ since $U_x$ is open. The boundary of $Z_x$ in $U_x$ (resp. in $S$) equals $Z_x$ deleting the interior of $Z_x$ in $U_x$ (resp. in $S$), so they agree too.
Can we show that $\stackrel{\circ}{Z_x}$ and ${Z_x}^c$ are the two connected components of $(\partial Z_x)^c$?
Manifolds are locally path-connected and the properties of being manifolds and being locally path-connected are both preserved by open subsets. For a locally path-connected space $X$, it is connected if and only if it is path-connected.
So $S$ is path-connected. By construction, $\stackrel{\circ}{Z_x}$ and $Z_x^c$ are both non-empty open in $S$. It suffices to show both of them are path-connected.
The former one is a homeomorphic image of an open disc which is clearly path-connected, so it is path-connected.
For the latter, pick distinct $y,z\in Z_x^c$, as $S$ is path-connected, there exists a continuous map $P:[0,1]\to S$ s.t. $P(0)=y,P(1)=z$. If $P^{-1}(Z_x)=\emptyset$ then $P$ factors through $Z_x^c\hookrightarrow S$ so it is a path joining $y$ and $z$ in $Z_x^c$.
If $P^{-1}(Z_x)\neq \emptyset$, then $P^{-1}(Z_x)$ is a proper closed subset of $[0,1]$ contained in $(0,1)$. Let $$s:=\inf{P^{-1}(Z_x)},t:=\sup {P^{-1}(Z_x)},$$ we must have $s\leq t$ in $P^{-1}(Z_x)$ by closureness of $P^{-1}(Z_x)$. In particular $s,t\notin \{0,1\}$. By continuity of $P$, there exists $\epsilon_s,\epsilon_t>0$ s.t. $$P((s-\epsilon_s,s+\epsilon_s)),P((t-\epsilon_t,t+\epsilon_t))\subset U_x$$ Hence $P(s-\frac{\epsilon_s}{2}),P(t+\frac{\epsilon_t}{2})\in U_x\backslash Z_x$. Clearly $U_x\backslash Z_x$ is a homeomorphic image of an open annulus in $\mathbb{R}^2$ so path-connected. There exists a path $L$ joinning $P(s-\frac{\epsilon_s}{2})$ and $P(t+\frac{\epsilon_t}{2})$ in $U_x\backslash Z_x$. Thus $$y=P(0)\stackrel{P}{\longrightarrow}P(s-\frac{\epsilon_s}{2})\stackrel{L}{\longrightarrow}P(t+\frac{\epsilon_t}{2})\stackrel{P}{\longrightarrow}P(1)=z$$ forms a path joining $y$ and $z$ in $Z_x^c$. Thus $Z_x^c$ is path-connected, the result follows.