It depends on how you see undirected edges in presence of directed edges. Depending on your need, you can have your own definition of 'strongly connected' and define it accordingly.
As far as I know, if one says 'directed graph' then one usually means that all edges are directed. And if a graph is not directed, then it is undirected.
Here, you can also treat undirected edges as 'bi-directed' edges i.e. you can traverse in any direction on these edges. If you see undirected edges this way then yes, you can call a graph which has at least one directed edge, a 'directed graph'.
I'm sorry, but your proof is incomplete; you fell into the so-called "induction trap".
For the induction step, you assume that any connected graph with $k$ vertices has at least $k-1$ edges; and you want to prove that any connected graph $G$ with $k+1$ vertices has at least $k$ edges. In trying to prove this, you assume that $G$ was obtained by adding a new vertex (which you confusingly call $k'$) to a connected graph with $k$ vertices. But you have not justified this assumption.
To justify your argument, you would have to show that every connected graph with $k+1$ vertices can be obtained by adding a vertex (and some edges) to some connected graph with $k$ vertices. In other words, you have to show that, given any connected graph with $k+1$ vertices, you can find a vertex whose deletion results in a connected graph. This is true, but requires a proof. (Well, it's true for finite graphs, which is what we're talking about. In a connected infinite graph, it's possible that deleting any vertex disconnects it.)
For an alternative approach, you can delete an arbitrary vertex $v$ from a $k$-vertex graph $G$, without worrying about whether $G-v$ is connected or not; you then apply the induction hypothesis to each connected component of $G-v$, however many there may be. In this approach, you have to use the style of induction where you prove the statement for $k$ assuming that it holds for all numbers smaller than $k$.
Best Answer
This graph is a tree and so it does not contain a cycle. So if there are two points $A,B$ with two paths between them $$P_1 : A\to ...\to B$$ and $$P_2 : A\to ...\to B$$ then path $P_2^{-1}\circ P_1$ would be a cycle wich starts and ends with $A$.