An exercise asks me to prove this statement. I, however, didn't use the fact $\mathrm G$ has $\mathrm T1$ axiom, and so it seems that this information is redundant. My proof goes as follows:
Assume $(g, h) \mapsto gh^{-1}$ is continuous. We can make a continuous map $G \to \mathrm{Hom}(G, G)$ by fixing the first argument, where $\mathrm{Hom}(G,G)$ denote continuous maps. So by fixing the first argument to be the neutral element $e$ we see that inverse $\iota$ is continuous. Now it suffices to take composition of $gh^{-1}$ and $\mathrm{id} \times \iota$ to see that the operation on $G$ is continuous. The other direction is the same. After we assume that both operation and inverse are continuous, we can compose operation with $\mathrm{id} \times \iota$ to see that $(g, h) \mapsto gh^{-1}$ is continuous.
Firstly, I'm afraid my argument might be incorrect. Moreover, I would very much like to see a proof that uses the fact that $\mathrm G$ has $\mathrm{T1}$ topology, so please help me. I suppose it should be used in the direction proving that $G$ is a topological group, but I don't really see how to use the $T1$ property.