Proof
(related to Theorem: Euler Characteristic for Planar Graphs)
 By hypothesis, $G$ is a connected planar graph with a planar drawing consisting of \(v\) vertices, \(e\) edges and \(f\) faces.
 We may assume that $v\ge 1,$ since a connected graph requires at least one vertex.
 First, we observe: If the formula $ve+f=2$ holds for a multigraph (i.e. an undirected graph with some loops and/or multiple edges), then it also holds for a simple graph obtained from this multigraph by removing all loops and replacing all multiple edges by simple edges.
 For if we remove a loop or remove in a given multiple edge one of the edges, the numbers $e$,$f$ and $v$ change as follows:
* $e\to e1$
* $f\to f1$
* $v\to v$
* an the formula remains unchanged.
 Therefore, we can assume that we have a connected simple graph, for which we have to check the formula $ve+f=2.$
 Case $v=1$:
 We have $e=0$ and $f=1$ (infinite face) and the formula is correct.
 Case $v=2$:
 We have $e=1$ and $f=1$ (infinite face) and the formula is again correct.
 Case $v > 2.$
 Since the graph is connected, we can remove edges one by one to get the case $v=2.$ When removing the edges, we have again two subcases:
* Case a) The toberemoved edge has one vertex with degree $=1$ and the other with degree $ > 1$.
* In this case, removing the edge would disconnect the graph unless we also remove the vertex with degree $=1$. Thus,
* $e\to e1$ (the removed edge)
* $f\to f$ (the surrounding face of the removed edge does not change).
* $v\to v1$ (the removed vertex with degree $=1$)
* Thus, the formula $ve+f=2$, if correct, does not change.
* Case b) Both vertices incident to the toberemoved edge have a degree $ > 1$.
* In this case, removing the edge will not disconnect the graph and we do not have to remove any vertices. Thus,
* $e\to e1$ (the removed edge),
* $f\to f1$ (two faces incident to the removed edge are melted together to one face),
* $v\to v$ (no removed vertices).
* Thus, the formula $ve+f=2$, if correct, does not change.
 Altogether, we have shown for the case $v > 2$ that we can transform the graph to the case $v=2,$ for which we have already proven the formula \[ve+f=2.\]
 It follows that the formula holds for all connected planar graphs with a planar drawing.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Aldous Joan M., Wilson Robin J.: "Graphs and Applications  An Introductory Approach", Springer, 2000