I am reading a paper about graphs. In this paper, authors wrote "we view infinite graph as cell complex with usual topology". All graphs are infinite here.
I know that we can consider graphs with following topology:
An arc $A_i$ is space homeomorphic to the unit interval [0,1].
Graph is a space $X$ that is the union of a collection of subspaces $A_i$ each of which is an arc, such that
(1) The intersection of two arcs $A_i\cap A_j$ is either empty or consists of a single point that is an end point of each.
(2) The toplogy of X is coherent with the subspaces $A_i$.
I am not familiar with cell complex and its topology. But, my question is Is the topology of cell complex as graph the same as with above topology?
Yes, they are the same. For a quick explanation, I'll paraphrase Hatcher's discussion from his book Algebraic Topology.
In terms of CW complexes, a graph is a space $X$ obtained from a discrete set $X^0$ be adding a collection of 1-cells $e_\alpha$.
In non-CW speak, this means $X$ is formed from the disjoint union of $X^0$ with closed intervals $I_\alpha$ by identifying the two endpoints of $I_\alpha$ with points in $X^0$. Here, the points in $X^0$ are the vertices of the graphs and the 1-cells $\{e_\alpha\}$ are the edges of $X$ (note that in this definition the edges does not include it's endpoints, so the closure $\overline{e_\alpha}$ of an edge $e_\alpha$ is the same as an arc $A_\alpha$ using your notation).
The usual topology is the quotient topology $X$ inherits from the disjoint union $$X^0 \coprod_\alpha I_\alpha$$ That is, a subset of $X$ is closed iff it intersects each $\overline{e_\alpha}$ in a closed set in $\overline{e_\alpha}$. So yes, the two definitions are equivalent.