Concrete proof of tightness of product measures on sequence space?

Consider the sequence space $\mathbb{R}^{\mathbb{Z}}$, which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.

A classical result on measure theory in Polish spaces tells us that any probability measure $\mu$ on a Polish space $X$ is tight: given $\epsilon > 0$, then there is a compact set $K_{\epsilon} \subseteq X$ such that $\mu(K_{\epsilon}) > 1 - \epsilon$.

Hence any product probability measure on $\mathbb{R}^{\mathbb{Z}}$ is tight.

Say we fix an i.i.d. probability measure $\mu$ on $\mathbb{R}^{\mathbb{Z}}$. Precisely, $\mu(\prod A_{i}) = \prod \nu(A_{i})$ for some probability measure $\nu$ on $\mathbb{R}$.

Is there a nice, concrete proof that $\mu$ is tight (independent of $\nu$)?

The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\nu$ has unbounded support, then, for each $M > 0$, the set $\{k \in \mathbb{Z} \, \mid \, |x_{k}| \geq M\}$ is infinite almost surely under $\mu$ (by the $0$-$1$ law). Thinking along those lines, it's not easy to see $K_{\epsilon}$ intuitively.

Here is one concrete example of such a set $K_\epsilon$. Let $M_n$ for $n\in \mathbb{N}$ be numbers such that $\nu([-M_n,M_n]^c) < \epsilon 2^{-n}$. Then, consider the set $$K_\epsilon = \{(a_n)_{n\in \mathbb{N}}\,:\, \forall n,\,|a_n|\le M_n\}\subset \mathbb{R}^{\mathbb{N}}.$$ We have $$\mu(K_\epsilon^c) \le \sum_{n = 1}^\infty \nu(\{|a_n|> M_n\}) < \epsilon\sum_{n = 1}^\infty 2^{-n} =\epsilon.$$ Also, $K_\epsilon = \prod_{n = 1}^\infty [-M_n,M_n]$ and so $K_\epsilon$ is the product of compact sets, and hence compact itself by Tychonoff's theorem.