If case of continuous random variables, we define median 'm' as a value such that the $P(X\ge m)=0.5$ and $P(X\le m)=0.5$. This link also mentions the same
For discrete random variables, $P(X\ge m)\ge0.5$ and $P(X\le m)\ge0.5$
Why is the cumulative probability required to be greater than or equal to $0.5$ in case of discrete random variables and only equals to $0.5$ in case of continuous random variable?
For strictly continuous random variables the value $m$ is unique and always in the support (the set of possible values) of $X$, thus there is no problem to define it as $P(X\ge m) = 0.5$. In discrete random variable whether with finite or infinite countable support, it is possible that no possible value of $X$ will satisfy $P(X\ge m )=0.5$. Let us consider a simple example. Let $X \sim \mathcal{B}in (3, 1/2)$, hence we have to find $m$ such that $ P(X \le m ) =0.5 $. Let us check possible value of $m$ $$ P(X\le 1) = \frac{1}{2^3} + \frac{3}{2^3} = 1/2. $$ Bingo! $1$ is the median. Now, take $p=1/4$ instead of $1/2$, $$ P(X\le 1) = \frac{3^3}{4^3} + \frac{3}{2^3} = 0.74, $$ that is way too high probability, however $$ P(X\le 0) = \frac{3^3}{4^3} = 0.42. $$ That is too low. I.e., there is no value that $X$ can get and it will satisfy $ P(X\le m) = 0.5, $ hence you have to chose $0$ or $1$ as the median. Any value in $(0,1)$ will not change a thing as $X$ cannot receive these values.