If 2 graphs are isomorphic, are they homeomorphic too? I'm quite confused about the homeomorphism definition. Vice-versa is definitely not true. But what can we say about the statement: 

If 2 graphs are isomorphic, they are homeomorphic .

 A: Two graphs $G_1,G_2$ are isomorphic, if there are mutually inverse morphisms of graphs $f:G_1\rightarrow G_2$ and $g:G_2 \rightarrow G_1$.
You can define a geometric/topological realization $\widehat{G}$ of a graph $G$ by taking the set of vertices and gluing copies of the unit interval $[0,1]$ to it wherever the graph has an edge. This construction is well behaved with respect to morphisms of graphs (aka functorial) in the sense that a morphism of graphs $f:G_1 \rightarrow G_2$ yields a continuous function $\widehat{f}:\widehat{G_1}\rightarrow \widehat{G_2}$ by sending some element $t\in [0,1]$ on an edge in $\widehat{G_1}$ connecting vertices $v$ and $w$ to $t\in[0,1]$ in $\widehat{G_2}$ of the edge connecting $f(v)$ and $f(w)$. One easily sees that this is continuous and likewise that this construction is well behaved with respect to composition.
Now one can manually check (or note that this is in fact implied by abstract nonsense) that an isomorphism of graphs is mapped under this construction to an isomorphism of topological spaces (aka homeomorphism).
The converse direction does not hold however. The problem is that the topological view can only see vertices of degree $\neq 2$. So for example the nonisomorphic graphs
$G_1 =\bullet -\bullet$ and $G_2=\bullet - \bullet - \bullet$ have homeomorphic topological realizations $\widehat{G_1} = [0,1]$ and $\widehat{G_2} = [0,2]\cong [0,1]$.
