Suppose $\alpha$ and $\beta$ are limit ordinals, and $f:\alpha\rightarrow\beta$ is strictly increasing and is cofinal in $\beta$, then $cf(\alpha)=cf(\beta)$
Well if such a function exists, then it must be the case that $\alpha\leq\beta$. Also if $\alpha=\beta$, then we are done. So suppose $\alpha<\beta$.
First I have to show that $cf(\alpha)\leq cf(\beta)$ by induction on $\alpha$:
If $\alpha=0$, then $cf(\alpha)=0<\omega\leq\ cf(\beta)$. So assume $cf(\alpha)<cf(\beta)$ for some limit ordinal $\alpha<\beta$. Let $\alpha'$ be a limit ordinal such that $\alpha<\alpha'<\beta$. It suffices to show $cf(\alpha')<cf(\beta)$. How can I use my inductive hypothesis, or any other assumptions I've made in the beginning?
There is no need to us an inductive argument. I'm not even sure that a reasonable inductive argument can be made. The thing to do here is to construct the right sequences, and remember the definition of $\operatorname{cf} ( \alpha )$: the least ordinal $\gamma$ such that there is a $\gamma$-sequence in $\alpha$ cofinal in $\alpha$.
Let $\gamma = \operatorname{cf} ( \alpha )$, and let $\delta = \operatorname{cf} ( \beta )$. We wish to show that $\gamma = \delta$, or, $\gamma \leq \delta$ and $\delta \leq \gamma$.
To show that $\gamma \leq \delta$ it suffices to find a $\delta$-sequence $( \alpha _ \xi )_{\xi < \delta}$ in $\alpha$ cofinal in $\alpha$. Start by choosing a $\delta$-sequence $( \beta_\xi )_{\xi < \delta}$ cofinal in $\beta$. For each $\xi < \delta$ since $f$ is cofinal in $\beta$ the set $\{ \mu < \alpha : \beta_\xi \leq f(\mu ) \}$ is nonempty, and has a minimal element; let $\alpha_\xi$ be the minimal element of this set. Given any $\mu < \alpha$ there is a $\xi < \delta$ such that $f(\mu) \leq \beta_\xi$, and by choice we have that $\beta_\xi \leq f( \alpha_\xi )$. As $f$ is strictly increasing it follows that $\mu \leq \beta_\xi$. Therefore $( \alpha_\xi )_{\xi < \delta}$ is cofinal in $\alpha$.
To show that $\delta \leq \gamma$ it suffices to find a $\gamma$-sequence $( \beta _ \xi )_{\xi < \gamma}$ in $\beta$ cofinal in $\beta$. Start by choosing a $\gamma$-sequence $( \alpha_\xi )_{\xi < \gamma}$ cofinal in $\alpha$. Consider the $\gamma$-sequence $( f(\alpha_\xi) )_{\xi < \gamma}$. Clearly this is a sequence in $\beta$. Given any $\nu < \beta$ as $f$ is cofinal in $\beta$ there is a $\mu < \alpha$ such that $\nu \leq f(\mu)$, and since $( \alpha_\xi )_{\xi < \gamma}$ is cofinal in $\alpha$ there is a $\xi < \gamma$ such that $\mu \leq \alpha_\xi$. Since $f$ is strictly increasing, we have that $\nu \leq f(\mu) \leq f(\alpha_\xi)$. Therefore $( f(\alpha_\xi) )_{\xi < \gamma}$ is cofinal in $\beta$.