If $H$ is a subgroup of a topological group $G$ and is equipped with the subspace topology, then $H$ is a topological group.

It's mostly a matter of seeing that the product topology $Top_{prod}$ on $H \times H$ matches the subspace topology $Top_{sub}$ on $H \times H$ coming from $G \times G$. The product topology on $H \times H$ is the smallest topology such that the two projections $\pi_1, \pi_2: H \times H \to H$ are continuous. An open in $H$ is of the form $U \cap H$ for some open $U$ of $G$, and $\pi_1^{-1}(U \cap H) = (U \cap H) \times H = (U \times G) \cap (H \times H)$ is open in the subspace topology on $H \times H$. A similar statement holds for $\pi_2$ in place of $\pi_1$. Hence $Top_{prod} \subseteq Top_{sub}$. But also, the subspace topology on $H \times H$ is the smallest topology such that the inclusion $i: H \times H \to G \times G$ is continuous. For opens $U, V$ of $G$ we have $i^{-1}(U \times V) = (U \times V) \cap (H \times H) = (U \cap H) \times (V \cap H)$; the latter is open in the product topology. So $Top_{sub} \subseteq Top_{prod}$.

If $m_H: H \times H \to H$ is the restriction of multiplication $m_G: G \times G \to G$, and $U \cap H$ is a typical open in $H$, then, $m_H^{-1}(U \cap H) = m_G^{-1}(U) \cap (H \times H)$ is open in $H \times H$ under the subspace topology, hence under the product topology by the above paragraph.

If $i_G: G \to G$ denotes inversion and $i_H: H \to H$ is its restriction, then $i_H^{-1}(U \cap H) = i_G^{-1}(U) \cap H$ is open in $H$ for a typical open $U \cap H$.