Prove $B_1 \subseteq B_2 \Rightarrow f^{-1}(B_1) \subseteq f^{-1}(B_2)$

Let $f : X \rightarrow Y$ and let $B \subseteq Y$. The inverse image of $B$ is a subset of $X$ defined by $$f^{-1}(B) = \{x \in X | f(x) \in B\}$$ Prove the following claims:

There are several claims that I am required to prove, but I am stuck on the first one, which is:

$$B_1 \subseteq B_2 \Rightarrow f^{-1}(B_1) \subseteq f^{-1}(B_2)$$

So far I have done:

$b \in (B_1 \subseteq B_2) \rightarrow b \in (f^{-1}(B_1) \subseteq f^{-1}(B_2))$

$\forall b [b \in B_1 \rightarrow b \in B_2]$ (Definition of a subset from my class' lecture notes.)

$\rightarrow f^{-1}(b) \subseteq f^{-1}(b)$

I feel like this is the right direction, but that I must have missed a step because it feels too "easy". Any advice?

Think of the function $f$ and the relation $\subseteq$ as 'arrows'. Then you just have to start at the correct place and 'follow the arrows' correctly:
Let $x$ be in $f^{-1}(B_1)$. Then by definition, $f(x)$ is in $B_1$. Since $B_1$ $\subseteq$ $B_2$, $f(x)$ is in $B_2$. Then again by definition, $x$ is in $f^{-1}(B_2)$. QED.
You want to show that if $x$ is an element of $f^{-1}(B_1)$, then $x$ must also be an element of $f^{-1}(B_2)$ (per the definition of subset). So suppose $x\in f^{-1}(B_1)$. Then there's some $b\in B_1$ so that $f(x)=b$. But since $B_1\subseteq B_2$, you know that $b$ is also an element of $B_2$. That is, we've found an element $b\in B_2$ so that $f(x)=b$. This is precisely the condition for $x$ to be an element of $f^{-1}(B_2)$.