Is the set of all finite sequences of letters of Latin alphabet countable/uncountable? How to prove either?

Proposition. If the alphabet $A$ is countable, the set $A^*$ of all finite strings in that alphabet is also countable.

Proof. $A, A^2 = A \times A, A^3 = A \times A \times A$, etc. are all countable and well-ordered (under the lexicographic ordering induced by the well-ordering of $A$), and $$A^* = \bigcup_{n=0}^{\infty} A^n$$ and the union of countably many well-ordered countable sets is again countable, so $A^*$ is countable. Note this does not require the axiom of (countable) choice.

However, the set $A^{\mathbb{N}}$ of all countably-infinite strings is countable if and only if $A$ is empty or is a 1-letter alphabet.

Suppose the 26 letters are "digits" in base-26. Any finite string of letters can be thought of as a unique positive integer in base-26. The positive integers are countable, so your set of strings must be as well.

(Actually, to be precise: letting the symbols in your set be non-zero digits establishes a bijection between your set and a subset of the natural numbers, so the cardinality of your set is less than or equal to that of the natural numbers; letting the symbols in your set be digits, including a "zero" digit, establishes a surjection from your set onto the natural numbers [leading zeros cause multiple digit strings to hit the same natural number], so the cardinality of your set is greater than or equal to that of the natural numbers; so since the cardinality of your set is ≥ and ≤ that of the natural numbers, they are equal in cardinality, so your set is countable.)