I am solving the following problem:
Let $(\Gamma_1, I)$ and $(\Gamma_2, I)$ be isomorphic graphs, then $|\Gamma_1|_I$ and $|\Gamma_2|_I$ are homeomorphic.
We denote by $(\Gamma, I)$ a graph, and its geometric realization by $|\Gamma|_I$.
The definitions is given in the lecture note Page 12, Definition: 2.2.1.
$\textbf{Added}$: A graph isomorphism $\phi= (\alpha, \beta)$ between two graph $\Gamma_1 = (V_1, E_1, i_1)$ and $\Gamma = (V_2, E_2, i_2)$ is a pair of bijective maps $\alpha: V_1 \rightarrow V_2$ and $\beta: E_1 \rightarrow E_2$ such that the following diagram commutes:
To solve the above problem, my approach is to prove that the following diagram commutes. 
Please help me.
The precise definition and contextual background to "geometric realisation" varies. But all you need to know is that geometric realisation is a functor. Two isomorphic graphs are related by morphisms of graphs: say, $f:\Gamma_1\to\Gamma_2$ is one such morphism. Then you have a continuous map $|f|:|\Gamma_1|\to|\Gamma_2|$.
If $g:\Gamma_2\to\Gamma_1$ is the inverse, so that $gf=\mathrm{Id}_{\Gamma_1}$ and $fg=\mathrm{Id}_{\Gamma_2}$, then $|g|:|\Gamma_2|\to|\Gamma_1|$ is continuous and we also know that $|g|\circ|f|=|gf|=|\mathrm{Id}_{\Gamma_1}|=\mathrm{Id}_{|\Gamma_1|}$ using the functorial property. Similarly, $|f|\circ|g|$ is also the identity.
So that's two continuous maps which are mutually inverse: the spaces $|\Gamma_1|$ and $|\Gamma_2|$ are then homeomorphic, by definition.
General fact: functors preserve isomorphisms.
Note: I have said $|\cdot|$ is a functor, but you have not provided a precise definition of your category of graphs. I just assume you are working with a sensible definition of that and of $|\cdot|$ so that we have the functorial property.
Definitions vary. One such definition is given here, and this definition is nice because our geometric realisation functor can be taken to be the standard one for simplicial sets.