Take the base field to be $\mathbb Q,$ set $\lambda=2^\omega=\mathfrak c,$ and $X_{\alpha}=\ell^\infty(\alpha)$: the bounded functions $\alpha\to\mathbb R.$ 
(If you prefer, you could use simple functions instead of all bounded functions.) The transition map $X_\alpha\to X_\beta$ is the restriction to $\beta.$

For each $\alpha>0,$ the space $X_\alpha$ has $\mathbb Q$-dimension $\mathfrak c.$ If $\dim X_0$ matters, take the example $X'_\alpha=X_{1+\alpha}$ instead.

For $\alpha<\beta<\omega_1$ the transition map has a kernel of $\mathbb R$-dimension at least one, so $\mathbb Q$-dimension $\mathfrak c.$

The limit $\varprojlim_{\delta<\gamma} X_\delta$ at countable limit ordinals $\gamma$ is the set of functions $\gamma\to\mathbb R$ that are bounded when restricted to $\delta$ for any $\delta<\gamma.$ This set contains $X_\gamma$ as a proper $\mathbb R$-subspace, so the $\mathbb Q$-dimension of $(\varprojlim_{\delta<\gamma} X_\delta)/X_\gamma$ is $\mathfrak c.$

The big limit $\varprojlim_{\alpha < \omega_1} X_\alpha$ is just $\ell^\infty(\omega_1)$ because any unbounded function on $\omega_1$ would be unbounded on some $\alpha<\omega_1.$ The dimension of this space is its cardinality, $2^{\omega_1}.$