Weak topology generated by identity maps

Let $(X, \tau_1)$ and $(X,\tau_2)$ be two topological spaces having the same underlying set X.
Let $\tau$ be the smallest topology on $X$ such that identity maps $I_1 : (X,\tau) \rightarrow (X,\tau_1)$ and $I_2 : (X,\tau) \rightarrow (X,\tau_2)$ are continuous.

Then if both $(X,\tau_1)$ and $(X,\tau_2)$ are separable topological space. Does it imply that $(X,\tau)$ will be separable?

It does not; here’s a counterexample.

Let $Y=\{y_k:k\in\Bbb N\}$ and $Z=\{z_k:k\in\Bbb N\}$ be disjoint countably infinite sets. Two subsets of $\Bbb N$ are said to be almost disjoint if their intersection is finite; let $\mathscr{D}$ be an uncountable family of almost disjoint subsets of $\Bbb N$. (Two constructions of such a family can be found in this answer.) Let $X=Y\cup Z\cup\mathscr{D}$. Let

$$\mathscr{I}=\big\{\{y_k\}:k\in\Bbb N\big\}\cup\big\{\{z_k:k\}\in\Bbb N\big\}\;.$$

For each $D\in\mathscr{D}$ let $D_Y=\{y_k:k\in D\}$ and $D_Z=\{z_k:k\in D\}$, and let

$$\mathscr{B}_1(D)=\{D\}\cup\{D_Y\setminus F:F\subseteq Y\text{ and }F\text{ is finite}\}$$

and

$$\mathscr{B}_2(D)=\{D\}\cup\{D_Z\setminus F:F\subseteq \text{ and }F\text{ is finite}\}\;.$$

Let

$$\mathscr{B}_1=\mathscr{I}\cup\bigcup_{D\in\mathscr{D}}\mathscr{B}_1(D)\qquad\text{and}\qquad\mathscr{B}_2=\mathscr{I}\cup\bigcup_{D\in\mathscr{D}}\mathscr{B}_2(D)\;;$$

$\mathscr{B}_1$ and $\mathscr{B}_2$ are bases for topologies $\tau_1$ and $\tau_2$, respectively, on $X$. Both topologies are separable: $Y\cup Z$ is dense in both.

The coarsest topology $\tau$ on $X$ making the identity maps from $\langle X,\tau\rangle$ to $\langle X,\tau_1\rangle$ and $\langle X,\tau_2\rangle$ continuous is the joint of $\tau_1$ and $\tau_2$, i.e., the topology generated by the subbase $\tau_1\cup\tau_2$, which has as a base the sets of the form $U\cap V$ with $U\in\tau_1$ and $V\in\tau_2$. Clearly $\mathscr{I}\subseteq\tau$. Let $D\in\mathscr{D}$ be arbitrary, and let $U=\{D\}\cup Y$ and $V=\{D\}\cup Z$; then $U\in\tau_1$ and $V\in\tau_2$, so $\{D\}=U\cap V\in\tau$. Thus, $\tau$ is the discrete topology on the uncountable space $X$, and $\langle X,\tau\rangle$ is not separable.