Cluster point and sub-sequential limit

I just have started to learn basic topology and came across a question in my class note: Suppose that $$U$$ is a compact Hausdorff topological space and let $$M$$ be a countable infinite subset of $$U$$. Let $$f:U\to V$$ be a continuous function, where $$V$$ is a metric space. We know that $$M$$ must have a cluster point say $$m$$ in $$U$$.

What I am confused about is that :

There must exists a sequence $$(u_n)$$ in $$M$$ such that $$f(u)$$ is a sub-sequential limit of $$(f(u_n))$$.

What happens if we replace the compact by locally compact? What is the function of $$U$$ being Hausdorff?

We use that $$U$$ is compact to show that $$M$$ has a cluster point. $$\mathbb{R}$$ with the standard topology is locally compact and Hausdorff, but if $$M=\mathbb{Z}\subseteq\mathbb{R}$$, then $$M$$ has no cluster point.
Of course, this doesn't directly apply to the part that you're confused about. But making $$U$$ compact is already a very strong condition! It implies, for example, that the range of $$f$$ is compact … so you should be able to take $$U$$ to be a compact metric space without loss of generality. Indeed, to make the above example more relevant, just let $$U=V$$, and take $$f$$ to be the identity.
• Thankyou @JacobManaker for your answer. Here $U$ is a general compact topological space! why we should take it metrizable?
• @user884919: My point is that you can replace $(f,U)$ with $(\text{id},f(U))$ in the problem, and then $f(U)$ is a subset of a metric space (and thus metric). Apr 11 '21 at 19:27