If $X$ is a compact metric space and $f$ is a continuous function from $X$ to real numbers, then $f$ is uniformly continous.
My question is is there any metric space such that every continous function is uniformly continous? If there exist such metric space can we conclude that $X$ is compact?
As already observed there are non-compact spaces like $\mathbb N$ where every continuous function is uniformly continuous. I will answer the question raised in the comments about the case of connected metric spaces.
Suppose every real continuous function on a connected metric space $X$ is uniformly continuous. We claim that $X$ is compact. In fact connectedness can be replaced by the weaker condition that $X$ has no isolated points.
Suppose $X$ has a sequence $\{x_{n}\}$ with no convergent subsequence. Choose positive numbers $\delta _{n},n=1,2,...$ such that $B(x_{n},\delta _{n})$ does not contain any $x_{m}$ with $m\neq n$ and $\delta _{n}<\frac{1}{% n}$ $\forall n.$ Since $x_{n}$ is not an isolated point for any $n$ we can find distinct points $x_{n,m},m=1,2,...$ in $B(x_{n},\delta _{n})\backslash \{x_{n}\}$ converging to $x_{n}$ as $% m\rightarrow \infty $ (for $n=1,2,...$). We may suppose $d(x_{n,m},x_{n})<% \frac{1}{n}$ $\forall n,m.$ Let $A=\{x_{n,n}:n\geq N\}$ and $B=\{x_{n}:n\geq 1\}$. Claim: $A$ is closed. Suppose $x_{n_{j},n_{j}}\rightarrow y$. Then $% d(y,x_{n_{j}})\leq d(y,x_{n_{j},n_{j}})+d(x_{n_{j},n_{j}},x_{n_{j}})<d(y,x_{n_{j},2m_{j}})+% \delta _{n_{j}}\rightarrow 0$. Thus $x_{n_{j}}\rightarrow y.$ Since $% \{x_{n}\}$ has no convergent subsequence it follows that $\{n_{j}\}$ is eventually constant and hence $y=\underset{j}{\lim }$ $x_{n_{j},n_{j}}$ belongs to $A.$ Clearly $B$ is also closed. Further, $A\cap B=\emptyset $. Let $f:X\rightarrow \lbrack 0,1]$ be a continuous function such that $% f(A)=\{0\}$ and$\ f(B)=\{1\}$. By hypothesis $f$ is uniformly continuous. Note that $d(A,B)\leq d(x_{n},x_{n,n})<\frac{1}{n}$ $\forall n.$ Thus $% d(A,B)=0$ and, by uniform continuity $d(f(A),f(B))=0.$ This contradicts the fact that $f(A)=\{0\}$ and$\ f(B)=\{1\}$. This contradiction shows that $X$ is compact.
