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Let $X=\{a,b,c,d,e\}$ be a space and $S=\Big\{\{a\},\{a,b,c\},\{b,c,d\},\{c,e\}\Big\}$ be a subbasis for a topology $T$ on $X$. How to find basis $B$ and $T$?

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    A basis $B$ for $T$ consists of all (finite) intersections of elements of $S$. The topology $T$ then consists of all unions of elements of $B$. The set $S$ is quite small, so this isn't too hard to compute. – 2017-01-28
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    Actually i am not getting how {a,b,c} {b,c,d} and {c,e} belong to B – 2017-01-28
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    $\{ a,b,c \} = \{ a,b,c \} \cap \{ a,b,c \}$ is an intersection of elements of the subbasis, hence it is an element of the basis. – 2017-01-28

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The base $B$ generated by the subbase is the set of all finite intersections from $S$. The intersection of $0$ members from $S$ yields (by convention) $X$ itself. All intersection of $1$ member from $S$ yields all members of $S$ themselves (So $S \subset B$ always). For 2 members we get $\{a\} \cap \{a,b,c\} =\{a\}$, nothing new. $\{a,b,c\} \cap \{b,c,d\} = \{b,c\} \in B$, we also get $\{c\} = \{b,c,d\} \cap \{c,e\} \in B$, the other intersections of two elements from $S$ yield no new members, so $$B = \{\emptyset, \{a\},\{a,b,c\},\{b,c,d\},\{b,c\}, \{c\} ,\{c,e\}, X\}$$.

Now the topology are all unions from $B$.

The topology is adequately descibed by just specifiying for each point $x$ its minimal open set $U_x$: for ours space these are

$$U_a = \{a\}, U_b =\{b,c\} ,U_c = \{c\} ,U_d = \{b,c,d\}, U_e = \{e\}$$

and $O \subseteq X$ is open iff $\forall x \in O: U_x \subseteq O$. So the open sets are just the different unions of the $U_x$.