Embeddings of an infinite graph. I'm asking for a formal proof or counterexample (if it's false) of the following.
If a graph (in general) has a vertex like in the picture below (of infinity degree), the
the graph cannot have a cellular embedding into any surface.
Any elaboration to help me build more intuition about this or similar questions
are very welcome.

 A: If a graph $G$ has a vertex $v$ of infinite valence then $v$ does not have a countable neighborhood basis.
To sketch a proof of this, let $E_i$ ($i \in I$) be the oriented edges of $G$ with initial vertex $G$, and let $\alpha_i : [0,1] \to E_i$ be an orientation preserving parameterization of $E_i$. For any function $R : I \to (0,1)$, the set $U(R) = \bigcup_{i \in I} \alpha_i \bigl([0,R(i))\bigr)$ is a neighborhood of $v$ (by definition of the celluar topology, also known as the CW topology). And then, for any countable set of neighborhoods $V_1,V_2,...$ of $f(v)$, one can choose a countably infinite subset $\{i_1,i_2,...\} \subset I$ and use a diagonalization construction to produce $R : I \to (0,1)$ so that $U(R)$ does not contain any of the subsets $V_j$.
But for any surface $S$ and any continuous injection $f : G \to S$, the image $f(v)$ does have a countable neighborhood basis in the subspace topology on $f(G)$, namely the intersection with $f(G)$ of any countable neighborhood basis of $f(v)$ in $S$. Therefore, $f$ is not an embedding, meaning that it does not have a continuous inverse with respect to the subspace topology on $f(G)$.
The point is that the cellular topology is a very fine topology, with many many open sets, too many for $G$ to be embeddable into a surface.
