How to show $f$ is uniformly continuous if it is continuous Let $X$ be a metric space such that every real-valued continuous function on $X$ is uniformly continuous on $X$. How to show that, given an arbitrary metric space $Y$ and a continuous function $f:X\to Y,$ $f$ is uniformly continuous on $X?$


Please help! I am just clueless to start.
A function $f:X\to {Y}$ is continuous at $a$ if and only if it is sequentially continuous at $a$.

Let $f$ is continuous at $a$. Let $\epsilon$ be given. Suppose $x_n\to {a}$ then there exists $\delta&gt;0$ such that $d(f(x),f(a))&lt; \epsilon$ for $d(x,a)&lt;\delta$ and there exists $M\epsilon \mathbb{N}$ such that $d(x_n,a)\delta$ for $n&gt;M$. It follows that $d(f(x_n),f(a))&lt;\epsilon$ for $n&gt;N$. So, $f(x_n)\to f(a)$ is sequentially continuous at $a$.

Conversely, suppose that $f$ is not continuous at $a$. Then there exists $\epsilon_0&gt;0$ such that for every $n\epsilon N$ there exists $x_n \epsilon X$ with $d(x_n,a)&lt;\frac{1}{n}$ and $d(f(x_n),f(a))&gt;\epsilon_0$ then $x_n\to a$ but $f(x_n)$ doesn't approaches $a$. So, $f$ is not sequentially continuous at $a$

Uniform continuity can be proved in the same way.
A function $f:X\to {Y}$ is continuous at $a$ if and only if it is sequentially continuous at $a$.

Let $f$ is continuous at $a$. Let $\epsilon$ be given. Suppose $x_n\to {a}$ then there exists $\delta&gt;0$ such that $d(f(x),f(a))&lt; \epsilon$ for $d(x,a)&lt;\delta$ and there exists $M\epsilon \mathbb{N}$ such that $d(x_n,a)\delta$ for $n&gt;M$. It follows that $d(f(x_n),f(a))&lt;\epsilon$ for $n&gt;N$. So, $f(x_n)\to f(a)$ is sequentially continuous at $a$.

Conversely, suppose that $f$ is not continuous at $a$. Then there exists $\epsilon_0&gt;0$ such that for every $n\epsilon N$ there exists $x_n \epsilon X$ with $d(x_n,a)&lt;\frac{1}{n}$ and $d(f(x_n),f(a))&gt;\epsilon_0$ then $x_n\to a$ but $f(x_n)$ doesn't approaches $a$. So, $f$ is not sequentially continuous at $a$

Uniform continuity can be proved in the same way.