Prove that a connected graph with $n$ vertices and $n+2$ edges is planar. What I have tried is that a tree has $n-1$ number of edges. So if there are $n$ edges there would be a cycle. This cycle would contribute to one face of the graph. Similarly for the other 3 edges. I think this would add one face each to the graph. Is this reasoning correct?

Let $G$ be a connected graph on $n$ vertices with $n+2$ edges.  Perform the following reductions on $G$.  
If a vertex $v$ of $G$ has degree $1$, then $G$ is planar iff $G-v$ is plannar.  Thus, we may remove all vertices of degree $1$ from $G$ successively, so that at the end, $G$ has no vertex of degree $1$.  
If $v$ is a vertex of degree $2$ of $G$, then let $u$ and $w$ be neighbors of $G$.  We remove $v$ from $G$ and add an edge $\{u,w\}$ to $G$.  Note that $u$ and $w$ may already be adjacent, so this procedure can result in a multigraph, but that is not a problem.  Indeed, if $u=w$ (i.e., $v$ is joined to $u$ by two edges), then this procedure removes $v$ and creates a loop at $u=w$.  This procedure does not change planarity of the graph $G$.
The reduction steps can be performed only fintely many times.  At the end, you will get a connected, possibly non-simple, graph $H$ each of whose vertices is of degree at least $3$.  If $m$ is the number of vertices of the graph $H$, then it is still the case that $H$ has $m+2$ edges.  However, by the Handshake Lemma,
$$2(m+2)\geq 3m\,,$$
and we conclude that $m\leq 4$.  It is easy to list all possible (not necessarily simple) connected graphs on $m$ vertices with $m\leq 4$ with $m+2$ edges, where the minimum degree is at least $3$, and check planarity for each of them.
The case $m=1$ has only $1$ nonisomorphic example.  The case $m=2$ has $4$ nonisomorphic examples.  The case $m=3$ has $5$ nonisomorphic examples.  The case $m=4$ has only $1$ nonisomorphic example.   (I hope I did not miscount.)
