Is every $T_1$ topological space homeomorphic to a subspace of a separable $T_1$ topological space (with the same cardinality)? Let $X$ be a $T_1$ topological space , then is it true that $X$ is homeomorphic to a subspace of a separable $T_1$ topological space with the same cardinality of $X$ ? If not then what if we drop the same cardinality requirement ? 
I know that the statement is true if we don't want the separable space t be $T_1$ , and I also know that the statement is false if we want the separable space to be $T_2$ because of cardinality restriction ( a separable $T_2$ space has cardinality at most $2^{c}$ ). But I can't figure out the $T_1$ case . Please help . Thanks in advance

We can define $\overline X = X \sqcup\{1,2,3,\ldots\}$ with the topology generated by these two flavours of open sets.


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*Sets of the form $U \sqcup \{n,n+1,\ldots\}$ for some $n \in \mathbb N$ and open $U \subset X$. 

*Sets of the form $\varnothing \sqcup W$ for any subset $W \subset \{1,2,3,\ldots\}$
Observe that, by construction, every open set of $\overline X$ has some $1,2,3,\ldots$ as an element. Therefore $\{1,2,3,\ldots\}$ is dense in $\overline X$
The singletons of the subspace $X \subset \overline X$ being closed follows from how $X$ is $T_1$.
We can separate any two elements $m,n \in \{1,2,3,\ldots\}$ using sets of the form $\varnothing \sqcup \{n\}$ and $\varnothing \sqcup \{m\}$.
We can separate any element $n \in \{1,2,3,\ldots\}$ and $x \in X$ using sets $\varnothing \sqcup \{n\}$ and $U \sqcup \{n+1,n+2, \ldots\}$.
