Compactness of a specific set in weak topology

I have the following question:

Let $$E$$ be a polish space (that is, a topological space, which is separable and metrizable, such that $$E$$ would be complete if equipped with this metric).

Consider the space of probability measures on $$E$$, denoted by $$\mathcal M (E)$$ equipped with the topology of weak convergence.

The claim I am struggling with is the following:

For every $$\ell\in\Bbb N$$ let $$K_\ell$$ be a compact set of $$E$$. Then the set $$\bigcap_{\ell\in\Bbb N} \{ m \in \mathcal M (E): m(K^{\operatorname{c}}_\ell) \leq \frac 1 \ell \}$$ is compact.

In this setting (i.e. $$E$$ polish) the set $$\mathcal M (E)$$ is metrizable with a complete metric, say $$d$$.
The set $$M := \cap_{\ell\in\Bbb N} \{ m : m(K_{\ell}^{\operatorname{c}}) \leq \frac 1 \ell \}$$ is tight (For any $$\varepsilon > 0$$ just choose $$\ell$$ such that $$\frac 1 \ell \leq \varepsilon$$ and take $$K_\ell$$). By Prokhorov this means that the closure of $$M$$ with respect to $$d$$ is compact.
Further, we have that for $$\eta >0$$ and $$K\subset E$$ compact the sets of the form $$\{ m : m(K^{\operatorname{c}}) \leq \eta \}$$ are closed with respect to $$d$$. This can be seen by the following: Take a convergent sequence sequence $$(m_n)_{n\geq 1}$$ with $$m_n (K^{\operatorname{c}}) \leq \eta$$ for every $$n$$. Define $$m := \lim_{n\to\infty} m_n$$. Since the topology induced by $$d$$ is equivalent to the topology of weak convergence we have the portmanteau theorem, such that: $$m(K^{\operatorname{c}}) \leq \liminf_{n\to\infty} m_n (K^{\operatorname{c}}) \leq \eta$$ All in all this means $$M$$ is compact.