Consider the set $\Sigma_m = \{0,1,2,,,m-1\}^{\mathbb{Z}}$ consisting of all two-sided sequences of elements in $\{0,1,2,...,m-1\}$. Let $T:\Sigma_m \to \Sigma_m $ be the left shift, which is a homeomorphism, i.e. $T(x_n)_{n\in \mathbb{Z}}=(x_{n+1})_{n\in \mathbb{Z}}.$
Let $A \subset \Sigma_m$ be a non-empty compact, closed and $T$-invariant set. Is it true that $T|_{A}:A \to A$ is subshift of finite type? By definition, it is subshift, but I don't know whether it is a subshift of finite type.
The answer is no; not all subshifts are of finite type.
As you declared a subshift on $m$ symbols is a closed shift invariant subset of $\Sigma_m$. One way to define subshifts is by declaring taboos. More precisely, denote by $W_m$ the set of all words in (i.e. tuples of) $m$ symbols (the set of which we fix as $\underline{m}=\{0,1,...,m-1\}$ throughout). Then for any subset $\mathcal{T}\subseteq W_m$, denote by $\Sigma_m(\mathcal{T})$ the subset of $\Sigma_m$ that consists of bi-infinite sequences that do not contain any word in $\mathcal{T}$. It's straightforward to see that $\Sigma_m(\mathcal{T})$ is a subshift for any subset $\mathcal{T}$ of words. Further, any subshift is of this form, so that we have a surjective map
$$\mathcal{P}(W_m)\to \operatorname{Sub}_m, \mathcal{T}\mapsto \Sigma_m(\mathcal{T}).$$
(Here the domain is the power set of $W_m$ and the codomain is the collection of all subshifts on $m$ symbols.)
Note that this map is not surjective (when $m\in\mathbb{Z}_{>1}$); that is, there may be multiple taboo collections that define the same subshift. A subshift $S\in\operatorname{Sub}_m$ is said to be of finite type if for some finite $\mathcal{T}\in\mathcal{P}(W_m)$, $S=\Sigma_m(\mathcal{T})$.
(I believe it's a nice exercise to work through how the above map works with set operations on the LHS.)
The standard example of a subshift that is not of finite type is the so-called even shift, due to B. Weiss (from his paper "Subshifts of Finite Type and Sofic Systems").
Ex (Even Shift): Put $m=2$ and $\mathcal{T}=\{1\,0\,1,1\,000\,1,1\,00000\,1,1\,0000000\,1,..., 1\, 0^{2n+1}\, 1,...\}$, that is, it's taboo to have an odd number of $0$'s between any two $1$'s.
Here is another example of a subshift that is not of finite type:
Ex (Context Free Shift): Put $m=3$ and $\mathcal{T}=\{0\, 1^\alpha\, 2^\beta\, 0\,|\, \alpha\neq\beta\}$, that is, it's taboo to have a block of $1$'s followed by a block of $2$'s between any two $0$'s, except when the two blocks are of the same length.
As a final remark I should mention that the two examples of subshifts not of finite type can be distinguished; the even shift is the next best thing after finite type; it's an example of a sofic shift (a sofic shift is by definition a factor of a subshift of finite type). The context free shift fails to be a sofic shift. Here is a related Venn diagram:
(I took this from Lind & Marcus' book An Introduction to Symbolic Dynamics and Coding (p.69), which is the standard reference for symbolic dynamics ("for its own sake", let's say).)
(For completeness the golden mean shift is defined by $m=2$ and $\mathcal{T}=\{11\}$.)