Let $X$ be a Banach space and $X^*$ its dual. Let $\{L_n\}$ be a sequence in $X^*$ such that $L_n \rightarrow L$ in the weak* sense, and suppose $\{x_n\}$ is a sequence in $X$ such that $x_n \rightarrow x$ in norm. The question I am working on asks if it is necessarily true that $L_n(x_n) \rightarrow L(x)$.
I believe the answer should be yes. Note that since $L \in X^*$, its norm $|L|$ is bounded. Let $\epsilon > 0$ be given. Then weak* convergence implies there exists $n_1 \in \mathbb{N}$ such that $$|L_n(x) - L(x)| < \epsilon/2 \quad \forall n \geq n_1.$$ Similarly, convergence in norm says there exists $n_2 \in \mathbb{N}$ such that $$\|x_n - x\| < \frac{\epsilon}{2|L|} \quad \forall n \geq n_2.$$ Let $N = \max\{n_1, n_2\}$, then for all $n \geq N$: $$|L_n(x_n) - L(x)| \leq |L_n(x_n) - L(x_n)| + |L(x_n - x)| \leq |L_n(x_n) - L(x_n)| + |L|\|x_n - x\| \\< \frac{\epsilon}{2} + |L|\frac{\epsilon}{2|L|} = \epsilon.$$
Is this correct or have I made any errors in my proof?
No, your proof is wrong. First, please don't write $|L|$ for the norm, write $\|L\|$. Next, a very common (but easily fixable) mistake when writing explicit $\epsilon$-$N$ proofs is the choice of $\frac{\epsilon}{2\|L\|}$; if $\|L\|=0$, then this quotient is meaningless. Instead use something like $\frac{\epsilon}{2\|L\|+1}$ (but as you'll see below, this isn't the right choice either... you need $\frac{\epsilon}{2M+1}$, where $M$ is defined below).
Anyway, those were trivial matters. The big issue is that you haven't justified why $|L_n(x_n)-L(x_n)|\to 0$ as $n\to\infty$. You say this can be made less than $\frac{\epsilon}{2}$, but that's not true based only on what you said. Look back at where $n_1$ comes from. This $n_1$ (and hence $N$) depends on the specific point $x$ in the Banach space. The correct way to argue is as follows: for each $n\in\Bbb{N}$, \begin{align} |L_n(x_n)-L(x)|&\leq |L_n(x_n)-L_n(x)|+|L_n(x)-L(x)|\\ &\leq \|L_n\|\cdot\|x_n-x\|+|L_n(x)-L(x)|\\ &\leq M\cdot\|x_n-x\|+|L_n(x)-L(x)|,\tag{$*$} \end{align} where $M:=\sup\limits_{k\in\Bbb{N}}\|L_k\|$. Note that since $L_k\to L$ in the weak* topology, the sequence of functionals $\{L_k\}_{k=1}^{\infty}$ is by definition pointwise convergent, hence pointwise bounded, so by Banach-Steinhaus theorem (i.e uniform boundedness principle), it follows that the sequence of operator norms $\{\|L_k\|\}_{k=1}^{\infty}$ is bounded as well, and thus the supremum, $M$, above is finite. Now, we can take the limit $n\to\infty$ on the RHS of $(*)$. Both terms vanish by hypothesis (the first due to norm convergence, and second due to weak* convergence); the key difference between my argument and yours is that in mine, the second term is $|L_n(x)-L(x)|$, the point of evaluation is simply $x$, not $x_n$. This proves $L_n(x_n)\to L(x)$.
If you want, you can now transcribe this into an $\epsilon$-$N$ proof, but you shouldn't need to.