Proof verification: $[L_1L_2:K] \leq [L_1:K][L_2:K]$

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Let $K \subseteq L_1,L_2 \subseteq L$ be fields. I want to show the following: $$[L_1L_2:K] \leq [L_1:K][L_2:K]$$


My attempt: The definition of $$L_1L_2:=L_1(L_2):=\bigcap \{F:F\text{ is a subfield of }L\text{ and }L_1 \subseteq F\text{ and }L_2 \subseteq F\}$$

$$\begin{split} [L_1L_2:K]&=\dim_K L_1L_2\\ [L_1:K]&=\dim_K L_1\\ [L_2:K]&=\dim_K L_2\\ \end{split}$$

I know that the $K$-vector space $L_1L_2$ is a linear subspace of the $K$-vector space $L$. The same is true for the $K$-vector spaces $L_1,L_2$.

From linear algebra, I know the following theorem: $$\dim(U_1+U_2)=\dim U_1 +\dim U_2 - \dim (U_1 \cap U_2)$$

If I can show that $L_1L_2= U_1+U_2$, the result should follow. Now I have to show that $L_1L_2=U_1+U_2$

Rewriting the definition of $$L_1L_2:=L_1(L_2):=\bigcap \{F:F\text{ is subfield of }L\text{ and }L_1 \cup L_2 \subseteq F\}$$

This means that $L_1 \cup L_2 \in L_1L_2$. From linear algebra, I know that $\operatorname{span}(U_1 \cup U_2)=U_1+U_2$. Thus, $L_1L_2=U_1+U_2$ and thus $[L_1L_2:K] \leq [L_1:K][L_2:K]$ holds.


My Question: Is my answer correct?