Consider a family of $K-$vector spaces $\prod_{i\leq n}K\to\prod_{i\leq n+1}K$ by embedding where embedding is done by identifying the basis $(1,0,\dots,0)\to (1,0,\dots,0,0),...$ and similarly for the rest.
$\textbf{Q:}$ Is $\lim\limits_n(\prod_{i\leq n} K)\cong\cup_i(\prod_{j\leq i}K)$ true? Should I conclude $\cup_i(\prod_{j\leq i}K)=\prod_N K$?
Note that LHS means $\prod_{j\in N}K$ whereas every element of RHS arises from a finite sequence. Do I even have an infinite non-trivial sequence in RHS but it seems that all my sequence on RHS must terminate at finite position? However, I can identify the limit in set theoretical sense and then use $lim_n(\prod_{i\leq n}K)=\sqcup_{n\in N}(\prod_{i\leq n}K)/\sim$ where $\sim$ identifies element by the canonical embedding. What is wrong with my thought process?
This answer is intended to sum up the discussion in the comments. With the intention of clearing up some misconceptions, a slightly more categorical analysis is given before arriving at the concrete example of the post.
Take $(V_i)_{i \geq 1}$ a sequence of vector spaces with $V_i \subset V_{i+1}$. This gives a diagram
$$ V_1 \subset V_2 \subset \dots V_n \subset \dots \tag{$\star$} $$
over the category $\omega = 1 \to 2 \to 3 \to \dots$ which can be thought of $\mathbb{N}$ as a poset: we have an arrow $n \to m$ iff $n \leq m$. Concretely, the diagram $(\star)$ is given by mapping $i$ to $Fi = V_i$ for each $i \geq 1$ and $n \to m$ which is $n \to n+1 \to \dots \to m$ as the composition of inclusions from each $V_i$ to the next, from $n$ to $m$. Now, having constructed a diagram, one may wonder whether $\lim F$ and $\operatorname{colim} F$ exist, and how they can be described.
Note that in the category $\omega$, we have an initial object (namely $1$). Is is an instructive exercise to prove that if a category $J$ has an initial object $i$ and $F : J \to \mathcal{C}$ is a diagram, then $\lim J = Fi$. Hence, for $(\star)$ we get
$$ \lim J = F(1) = V_1. $$
As for the colimit of $(\star)$, I claim we have $\operatorname{colim} F \simeq \bigcup_{i \geq 1}V_i$. It suffices to see that the latter verifies the universal property of the colimit: we have inclusion maps $j_r : V_r \to \bigcup_{i \geq 1}V_i$ for each $s$ which commute with the inclusions $V_i \subset \dots \subset V_{j}$. It suffices to check that they commute for $V_i \subset V_{i+1}$ (why?). Noting the inclusions $ s_k : V_k \hookrightarrow V_{k+1}$ we get that $j_{k+1}s_k(v) = j_{k}(v)$ for each $v \in V_k$. This is because the first expression is looking at $v$ as an element of $V_{k+1}$, and then embedding it in the union, and the latter is directly looking at $v$ as a vector on the union. Hence we have built a cone $(j_r : V_r \to \bigcup_{i \geq 1} V_i)_{r \geq 1}$ under $F$. Finally, let's prove that it is universal in this sense: if we have another cone $(\mu_r : V_r \to W)_{r \geq 1}$, then the function
$$ \mu : v \in V_r \subset \bigcup_{i \geq 1}V_i \longmapsto \mu_r(v) \in W $$
is well defined, linear and the only one such that the new cone factorizes via the original inclusions, i.e. $\mu$ is the only linear function such that $\mu j_r = \mu_r$. This completes the proof of $\operatorname{colim} F \simeq \bigcup_{i \geq 1} V_i$.
Now, going back to the original question: here $V_i = \mathbb{k}^i$ and so $\lim F = \varinjlim V_i = \mathbb{k}$ and $\operatorname{colim} F = \varprojlim V_i = \bigcup_{i \geq 1} \mathbb{k}^i$. Hence if we interpret the limit as an inverse limit (i.e. a colimit) the statement is true. What is false is that $\bigcup_{i \geq 1} \mathbb{k}^i \simeq \mathbb{k}^{\mathbb{N}}$. The former has countable dimension, a basis given by $e_i = (0, \dots, 0, \overbrace{1}^i,0,\dots)$, but the latter does not.