Let $E$ and $F$ be topological vector spaces and $A \subset E$. I want to prove that: if $f: A \longrightarrow F$ is a uniformly continuous function, then $f$ is continuous.
I want that, by definition of uniformly continuous we have for all $V \in \mathcal{F}(0_F)$ there exists $U \in \mathcal{F}(0_E)$ such that if $x_1,x_2 \in A$ and $x_1-x_2 \in U$ implies $$f(x_1)-f(x_2) \in V$$ where $0_E$ and $0_F$ denote the origin of $E$ and $F$ respectively. Moreover $ \mathcal{F}(0_E)$ and $ \mathcal{F}(0_F)$ are the filters of neighborhoods of the origins.
But, I am not able to tie these facts with continuity, since the definition of continuity is via $f$ inverse images.
Fix $x\in A$ and let $V$ be an open set around $f(x)$. Then $-f(x)+V\in\mathcal{F}(0_F)$, so there exists $U\in\mathcal{F}(0_E)$ such that $x_1,x_2\in A$ with $x_1-x_2\in U$ implies $f(x_1)-f(x_2)\in -f(x)+V$. Consequently, $(x+U)\cap A$ is an open set in $A$ containing $x$ such that $f((x+U)\cap A)\subset V$, so $f$ is continuous at $x$ for all $x\in A$.
Edit: For clarity, the notion of continuity I used here is that a function between topological spaces $f:X\to Y$ is continuous if and only if for every $x\in X$ and every open set $V$ in $Y$ containing $f(x)$, there exists an open set $U$ in $X$ containing $x$ such that $f(U)\subset V$. It is a standard exercise in topology that this is equivalent to other definitions of continuous.