Supppose $ f : X \rightarrow Y $ is a function and $ A \subseteq X $ is countable. Then $ f(A) $ is countable. Supppose $ f : X \rightarrow Y $ is a function and $ A \subseteq X $ is countable. Then $ f(A) $ is countable. 
I am in an introduction to proofs course... we are studying topology in order to use that to write proofs. I have had a very brief introduction to countable sets and must show this statement is either true and prove it or show that it is false and give a counter example.
For proving a set is countable I know you need to show it is finite or countably infinite but I am still vague on how to do that... 
 A: The statement is not entirely accurate. For instance, suppose $A\subseteq X$ countable and we define $f(x):=y$ with $y\in Y$ fix.
So, the statement should be 

Let $f\colon X\to Y$ be a function, and let $A\subseteq X$ be a countable set. Then $f(A)$ is at most countable. (At most countable is equivalent to finite or countable.)

You can prove before the follow statements:
Proposition 1. Let $Y$ be a set, and let $f\colon\mathbf N\to Y$ be a function. Then $f(\mathbf N)$ is at most countable.
Corollary 2. Let $X$ be a countable set, and let $f\colon X\to Y$ be a function. Then $f(X)$ is at most countable.
Now, you can prove your statement by Corollary 2 and the restriccion $f|_A$.
Hint. In the Proposition 1, the basic problem is that is not assumed to be one-to-one. Define $A$ to be the set $$A:=\{n\in\mathbf N:f(m)\ne f(n)\;\text{for all}\;0\le m\le n\};$$ informally speaking, $A$ is the set of natural numbers $n$ for which $f(n)$ does not appear in the sequence $f(0),f(1),\dotsc,f(n-1)$. Prove that when $f$ is restricted to $A$, it becomes a bijection from $A$ to $f(\mathbf N)$. The use the proposition
Proposition 0. Let $X$ be an infinite subset of the natural numbers $\mathbf N$. Then there exists a unique bijection $f\colon\mathbf N\to X$ which is increasing, in the sense $f(n+1)>f(n)$ for all $n\in\mathbf N$. In particular, $X$ has equal cardinality with $\mathbf N$ and is hence countable.
A: Suppose $A$ is countable. We can number the elements of $A$, and call them $x_n$. We have
$$A = \left\lbrace x_n : n\in \mathbb{N} \right\rbrace $$
Then
$$f(A) = \bigcup_{n\in\mathbb{N}} \left\lbrace f(x_n) \right\rbrace = \left\lbrace y_n = f(x_n) : n\in\mathbb{N} \right\rbrace  $$
And $n\mapsto y_n$ is then obvisoulsly a surjection from $\mathbb{N}\to f(A)$
A: Use the following definition for countability:

A non empty set $A$ is countable if there is a surjection $\varphi:\Bbb N\to A$.

Suppose that $A$ is countable and $\varphi:\Bbb N\to A$ is the surjection. Define a map $\psi:A\to f(A)$ by $\psi(x)=f(x)$. Then clearly $\psi$ is a surjection. Now define $g:\Bbb N\to f(A)$ by $g=\psi\circ\varphi$. Since composition of two surjection is again a surjection, it follows that $g$ is a surjection and hence $f(A)$ is countable.
A: Let $A = \{ x_n \}_n$. Then you have $f(A) = \{ f(x_n) \}_n$.
Define $\phi: f(A) \to \mathbb{N}$ by $\phi(f(x_n)) = \min \{ k | f(x_k) = f(x_n) \}$. It is easy to check that $\phi$ is injective.
Let $I = \phi(f(A)) \subset \mathbb{N}$.
If $I$ is finite, you are finished, otherwise $I$ is infinite. Since $I \subset \mathbb{N}$, we can order the elements in increasing order to get
$I = \{k_1,k_2,... \}$. Note that the map $\eta:\mathbb{N} \to I$ given by $\eta(i) = k_i$ is a bijection.
Then $\eta^{-1} \circ \phi$ is a bijection between $f(A)$ and $\mathbb{N}$.
A: For $f$ to be a function, each $x \in A$ can get mapped to at most one value in the range of $f$. For each $y$ in the range of $f$, $y$ corresponds to at least one $x \in A$. You should be able to infer (or easily prove) from this realization that $|f(A)| \leq |A|.$ So $f(A)$. 
As for you being vague on the nature of countability, here is some more info. A set $C$ is countable iff it has a one-to-one correspondence with a subset of the natural numbers. That is, for some $N \subset \Bbb{N}$ there exists a bijection $g:N \to C$ such that $g(n) = c_n \in C$ for each $n \in N$. The entire definition of countable is most often written in the condensed form of $C = \{c_1,c_2,c_3,\ldots \}$ or $C = \{c_i\}_{i \in \Bbb{N}}$. However, under this definition of countable, sets that are finite or countably infinite are all considered countable, since $N$ doesn't have to be a proper subset of $\Bbb{N}$. However, it is often the case that when one is given "a countable set" in a theorem, it is implied to be countably infinite, and not finite. So the definition is a bit ambiguous in practice. 
Anyway, with the proof that you are given, you are allowed to assume that $A$ is numerable, so you can write $A = \{a_1,a_2, a_3, \ldots \}$. You want to show that $f(A)$ can be written as a numerable set in the form $\{f(a_1),f(a_2),f(a_3)\ldots \}$. See if you can combine the definition of countable with what I said in my first paragraph, and perhaps use some of the things other users have answered with. It might help if for practice you first prove 

If $A\subset X$ is finite and $f$ is a function from $X \to Y$ then $f(A)$ is finite.

Once you prove this, you may have an easier time proving this result for when $A$ is countably infinite.
