Problem :
Let $X,Y,Z \subset \mathbb R$ and assume that $f:X\to Y$ is uniformly continuous on $X$ and $g:Y\to Z$ is uniformly continuous on $Y$. Then $(g \circ f):X \to Z$ is uniformly continuous on $X$.
This is really easy but i want to clarify one thing, $\forall \epsilon >0 ~\exists ~ \delta >0 : \forall x_0,x_1 \in X$, $|x_0-x_1|<\delta \implies |f(x_0)-f(x_1)| < \epsilon$.
Fix a $\delta_1 \leq \epsilon$(appropriate one to go according to uniform continuity) and thus $\forall y_1,y_2 \in Y$ $|y_1-y_2|<\delta_1 \implies |g(y_1)-g(y_2)|<\epsilon_1 ~ \forall~ \epsilon_1>0$.
And here's my doubt, we do not know that $X,Y$ is closed, so we cannot guarantee surjectivity of $f,g$ and thus we cannot guarantee that $\exists x_3,x_4 \in X ~: ~ f(x_3)=y_1,f(x_4)=y_2$ (so that we can place this values in the secind expression to obtain composition) and this is troubling me, and i have a little confusion here.
I woukd be really grateful if you help me here, many thanks.
Surjectivity is not needed. You take a $\delta^*$ such that$$\lvert y_0-y_1\rvert<\delta^*\implies\bigl\lvert g(y_0)-g(y_1)\bigr\rvert<\varepsilon.$$Then you take a $\delta>0$ such that$$\lvert x_0-x_1\rvert<\delta\implies\bigl\lvert f(x_0)-f(x_1)\bigr\rvert<\delta^*.$$And then, if $\lvert x_0-x_1\rvert<\delta$, then $\bigl\lvert f(x_0)-f(x_1)\bigr\rvert<\delta^*$ and therefore $\bigl\lvert g\bigl(f(x_0)\bigr)-g\bigl(f(x_1)\bigr)\bigr\rvert<\varepsilon$.