It is indeed. The price of an American option is the Bermuda option in the limit that the exercising interval approaches zero. The Bermuda option at any exercising time can be evaluated inductively via the dynamic programming principle as the maximum of the payoff and the risk-neutral expected value of the Bermuda option price at the next exercise time. The latter is inductively assumed to be whilst the former is convex in the strike. The maximum of convex functions is again convex. The dominant convergence theorem guarantees the pointwise limit of a sequence of convex functions is again convex. Therefore the American option is convex in strike. As a matter of fact the same deduction applies to an option where the principle of dynamic programming is applicable any the payoff function is convex with respect to a parameter.

We will show the convergence of the Bermuda option price to its associated American option price. 

Let $A$ be the price at time $0$ of an American option with a continuous payoff function $g(S)$ on the underlying $S$ expiring at time $1$, i.e.
$$A=\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_\tau),$$
where $\mathbb FS$ stands for the set of all stopping times taking value in set S. Let $T_n:=\{0,t_1,t_2,\cdots,t_{n-1},t_n=1\}$, where $0<t_1<t_2<\cdots<t_{n-1}<1$ and $\max_{0\le i\le n-1}(t_{i+1}-t_i)\to 0$ as $n\to\infty$. The associated Bermuda option price at time $0$ is
$$B_n=\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau).$$

Define [simple function][1]
$$\tau_{T_n}:=\sum_{i=0}^{n-1} t_i\mathbf 1_{[t_i,t_{i+1})}.$$
$\tau_{T_n}\to \tau$ almost surely as $n\to\infty$. 

**Lemma:** $$\lim_{n\to\infty} B_n=A.$$

**Proof:** Fix an arbitrary stopping time $\tau\in\mathbb F[0,1]$ and $\epsilon>0$.

Since $g$ is continuous, by the dominated convergence theorem,
$$\mathbf E|g(S_{\tau_{T_n}})-g(S_{\tau})|\to0$$
as $n\to\infty$. $\exists N(\tau,\epsilon)\ni$
$$\mathbf Eg(S_{\tau_{T_n}})>g(S_\tau)-\epsilon \tag1$$
for $\forall n>N$. For such $n$
$$B_n=\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau)\ge Eg(S_{\tau_{T_n}}) \tag2.$$
There are infinitely many $n>N$, that 
$$\liminf_{n\to\infty} B_n\ge B_n \tag3$$
Combining $(1), (2)$ and $(3)$, we have 
$$\liminf_{n\to\infty} B_n>\mathbf Eg(S_{\tau})-\epsilon.$$

As $\tau$ and $\epsilon$ are arbitrary
$$\liminf_{n\to\infty} B_n\ge \sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau}).$$

Now it is obvious
$$\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau})\ge\sup_{\tau\in\mathbb FT_n}\mathbf Eg(S_\tau)=:B_n.$$
That leads to
$$A=\sup_{\tau\in\mathbb F[0,1]}\mathbf E g(S_{\tau})\ge\liminf_{n\to\infty} B_n. \tag4$$

Finally, combining $(3)$ and $(4)$, we conclude that
$$\lim_{n\to\infty} B_n=A.$$


  [1]: https://en.wikipedia.org/wiki/Simple_function