Proposition 2

Let $X, X', X''$ be three topological spaces; let $f$ be a mapping of $X$ into $X'$, continuous at $x \in X$; let $g$ be a mapping of $X'$ into $X''$, continuous at $f(x)$. Then the composition $h = g \circ f : X \to X''$ is continuous at $x$.

Let $V''$ be a neighbourhood of $h(x) = g(f(x))$ in $X''$. Since $g$ is continuous at $f(x)$ it follows that $\overline{g}^{-1}(V'')$ is a neighbourhood of $f(x)$ in $X'$. But $f$ is continuous at $x$, hence $\overline{f}^{-1}(\overline{g}^{-1}(V'')) = \overline{h}^{-1}(V'')$ is a neighbourhood of $x$ in $X$, and therefore $h$ is continuous at $x$.