Sequential Compactness is Preserved under Continuous Surjection

Theorem
Let $T_A = \left({X_A, \vartheta_A}\right)$ and $T_B = \left({X_B, \vartheta_B}\right)$ be topological spaces.

Let $\phi: T_A \to T_B$ be a continuous surjection.

If $T_A$ is sequentially compact, then $T_B$ is also sequentially compact.

Proof
Let $T_A$ be a sequentially compact space.

Take an infinite sequence $\{x_n\}\subset X_B$.

From the surjectivity of $\phi$ ,there exists another infinite sequence $\{y_n\}\subset X_A$ such that $\phi\left(y_n\right)=x_n$.

By the definition of sequential compactness, there exists a subsequence of $\{y_n\}$, let's call it $\{y_{n_k}\}$, such that it converges to $y\in X_A$ with respect to $T_A$.

From the continuity of $\phi$, it is concluded that $\phi(y_{n_k})=x_{n_k}$ converges to $\phi(y)\in X_B$.

Thus, $\{x_n\}$ has a subsequence that converges; so $T_B$ is sequentially compact by definition