existence of a specific topology on $\mathbb{R}$ Is there a topology on $\mathbb{R}$ such that for every subbasis of it like $S$ 
,we have $Card(S) > Card(\mathbb{R})$ ?
 A: Equivalent reformulation: is there a topological space $X$ of size $\mathfrak{c}$ and  $w(X) > \mathfrak{c}$. This holds because the weight is the minimal size of a base for $X$, which is also the size of a minimal subbase for $X$, as finite intersections from subbase elements from an infinite family do not raise the cardinality. Then transport the topology to $\mathbb{R}$ via your favourite bijection.. 
One way to find such a space is to take an ultrafilter $\mathscr{U}$ on a set $X$. Then define  $\mathcal{T} = \mathscr{U} \cup \{\emptyset\}$, which is a topology (we added $\emptyset$, $X$ is in any filter, filters are closed under finite intersections, and being filters closed under enlargements trivially implies that $\mathcal{T}$ is closed under unions, so even for normal filters $\mathscr{F}$ we get a topology this way; note the cofinite topology is a well-known example of this construction, as is the co-countable topology). For an ultrafilter, this ultrafilter topology has no base (and so subbase) of size $\le |X|)$. I will assume that $\mathscr{U}$ is uniform, i.e. for every $O \in \mathscr{U}$. $|O| = |X|$. These can be "constructed" by extending the family $\{A: |X\setminus A| < |A|\}$, which has the FIP ,to an ultrafilter by Zorn.
Proof: Suppose that $\{B_\alpha, \alpha < |X|\}$ is a base for $\mathcal{T}$, in particular all of them are members of $\mathscr{U}$ such that any $O \in \mathscr{U}$ is a union of a subfamily of them. Now also enumerate $X = \{x_\alpha : \alpha < |X|\}$ and define two sequences $p_\alpha, q_\alpha$
as follows, for all $\alpha < \beta < |X|$:


*

*$\forall \alpha: p_\alpha \in B_\alpha, q_\alpha \in B_\alpha$

*$p_\beta \notin \{p_\alpha, q_\alpha : \alpha < \beta\}$

*$q_\beta \notin \{p_\alpha, q_\alpha : \alpha < \beta\}$


This can be done as at any stage $\beta$ the set of already chosen points is of size at most $\beta$ and $|\beta| < |X|$ and every $B_\beta$  has size $|X|$, this is where uniformity of the filter is handy.
Now define $P = \{p_\alpha: \alpha < |X|\}$, and  $Q = \{q_\alpha: \alpha < |X|\}$ and note that by construction $P \cap Q = \emptyset$
Now $P \in \mathscr{U}$ or $X \setminus P \in \mathscr{U}$ as we have an ultrafilter. 
Suppose $P$ is, by assumption, we have a non-empty subset $I$ of indices, such that  $P = \cup\{B_\alpha : \alpha \in I\}$. But then $P$ would contain any $q_\beta$ for $\beta \in I$, as $q_\beta \in B_\beta$, which cannot be.
We get a similar contradiction assuming that $X \setminus P$ is such a union, find a $p_\beta$ in it.. So the  $U_\alpha$ cannot form a base, and this shows that any (sub)base for $\mathcal{T}$ has size strictly more than $X$.
So the eventual answer is yes: take any uniform ultrafilter on $\mathbb{R}$ and use it as a topology.  
Here are some notes that prove the claim about uniform ultrafilters, and in general contains interesting info on ultrafilters. 
