I have to prove that if $f:X\to Y$ and $g:Y\to Z$ are continuous applications between metric spaces, then $g\circ f$ is continuous.
I prove this statement only for a point $x_0\in X$ (it's sufficient).
Let $x_0\in X$.
Because $f$ is continuous in $x_0$,
$\forall \varepsilon>0,\quad \exists \delta_f>0$ such that $ d_X(x,x_0)<\delta_f \Longrightarrow d_Y\left(f(x),f(x_0)\right)<\varepsilon$
but $g$ is continuous in $y_0=f(x_0)$, so,
$\exists \delta_g>0$ such that $d_Y(y,y_0)<\delta_g \Longrightarrow d_Z\left(g(y),g(y_0)\right)<\varepsilon$
Let's consider $g\circ f :X\to Z$ and define $\delta=\min\{\delta_f,\delta_g\}$.
So, because $\delta\leq \delta_f$, we have:
$\forall \varepsilon>0,\quad d_X(x,x_0)<\delta \Longrightarrow d_Y\left(f(x),f(x_0)\right)<\varepsilon$
and, because $\delta\leq \delta_g$, we have:
$\forall \varepsilon>0, \quad d_Y(y,y_0)<\delta \Longrightarrow d_Z\left(g(y),g(y_0)\right)<\varepsilon$
So, $\forall \varepsilon>0,\quad \exists \delta>0$ such that $ d_X(x,x_0)<\delta \Longrightarrow d_Z\left(g\left(f(x)\right), g\left(f(x_0)\right)\right)<\varepsilon$. Then, $g\circ f$ is continuous in $x_0$.
Does it works?