Let $X$ be a space endowed with two topologies $\tau, \tau'$. Then one can show that $\tau\cap \tau'$ is also a topology on $X$, where $$ \tau \cap \tau' = \{ t \cap t' : t\in \tau, \ \ t'\in \tau'\}.$$
This intersection is the greatest lower bound in the partially ordered set of all topologies on $X$. The answer below shows that $\tau\cap\tau'$ need not be Hausdorff even if $\tau$ and $\tau'$ are generated by metrics.
Now assume that $\tau, \tau'$ are both semi-metrizable, i.e., generated by semi-metrics $d$ and $d'$ (which are the same as metrics but without the requirement that $d(x,y)=0$ implies $x=y$).
Is it then true that $\tau\cap \tau'$ is also semi-metrizable? If it is true, are there explicit relationship between $d, d'$ and the new metric?
The $\tau\cap\tau'$-neighbourhoods of a point $x$ are precisely those sets which contain the union $B_d(x,r)\cup B_{d'}(x,r)$ for some $r>0$ which suggests to define $\tilde d(x,y)=\min\{d(x,y),d'(x,y)\}$ which, however, does not satisfy the triangle inequality and thus fails to be a metric.
Here’s an example to show that $\langle X,\tau\cap\tau'\rangle$ need not even be Hausdorff.
Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are any two distinct points not in $\Bbb N$. Let
$$\tau=\wp\big(X\setminus\{p\}\big)\cup\{U\subseteq X:p\in U\text{ and }\Bbb N\setminus U\text{ is finite}\}$$
and
$$\tau'=\wp\big(X\setminus\{q\}\big)\cup\{U\subseteq X:q\in U\text{ and }\Bbb N\setminus U\text{ is finite}\}\;.$$
In other words, in $\langle X,\tau\rangle$ every point except $p$ is isolated, and nbhds of $p$ are cofinite. Replace $p$ by $q$ to get $\tau'$. I’ll leave it to you to check that $\langle X,\tau\rangle$ and $\langle X,\tau'\rangle$ are both homeomorphic to the subspace
$$\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$
of the real line with the usual topology and are therefore metrizable.
Now suppose that $p\in U\in\tau\cap\tau'$ and $q\in V\in\tau\cap\tau'$. Since $p\in U\in\tau$, we know that $\Bbb N\setminus U$ is finite, and since $q\in V\in\tau'$, we also know that $\Bbb N\setminus V$ is finite. From this it’s easy to deduce that $U\cap V\ne\varnothing$.