Sequence of convex functions converges uniformly

Sequence of convex functions converges uniformly

I am working on the following problem.
Let $f_{n}: [a, b] \rightarrow \mathbb{R}$ be a sequence of convex
functions. Furthermore, for each fixed $x \in [a, b]$, suppose $f(x) =
\lim_{n \rightarrow \infty}f_{n}(x)$ exists and $f(x)$ is continuous on
$[a, b]$. Show that $f_{n} \rightarrow f$ uniformly.
The solution seems to have been mentioned here and here. However, the
solutions don't seem very satisfactory to me, so I decided to write out
explicitly my own solution (partially because they start with a proof by
contradiction). Can anyone check my solution to make sure it is correct?
I've tried to formulate a direct proof.
Since $f$ is continuous on $[a, b]$, it is uniformly continuous there, and
hence there exists an $\delta > 0$ such that when $|x - y| < \delta$,
$|f(x) - f(y)| <\varepsilon/10$.
Partition $[a, b]$ into $\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{m}\}$
where $\alpha_{0} = a$ and $\alpha_{m} = b$ such that $|\alpha_{i + 1} -
\alpha_{i}| < \delta/2$, $i = 0, 1, 2, \ldots, m - 1$.
Since for each fixed $x$, $f_{n}(x)$ converges to $f(x)$, there exists an
integer $N$ such that $|f_{n}(\alpha_{i}) - f(\alpha_{i})| <
\varepsilon/10$ for all $i = 0, 1, \ldots, m$ and $n \geq N$.
For $\bar{x} \in [a, b]$, $\bar{x} \in [\alpha_{i}, \alpha_{i + 1}]$ for
some fixed $i$. Then $\bar{x} = \lambda_{\bar{x}} \alpha_{i} + (1 -
\lambda_{\bar{x}})\alpha_{i + 1}$ for $\lambda_{\bar{x}}$ between 0 and 1.
Therefore by convexity of $f_{n}$, for $n \geq N$ (which is independent of
$\bar{x}$), we have \begin{align*} |f_{n}(\bar{x}) - f(\bar{x})| &\leq
|\lambda_{\bar{x}} (f_{n}(\alpha_{i}) - f(\bar{x})) + (1 -
\lambda_{\bar{x}})(f_{n}(\alpha_{i + 1}) - f(\bar{x}))|\\ &\leq
|f_{n}(\alpha_{i}) - f(\bar{x})| + |f_{n}(\alpha_{i + 1}) - f(\bar{x})|\\
&\leq |f_{n}(\alpha_{i}) - f(\alpha_{i})| + |f(\alpha_{i}) - f(\bar{x})| +
|f_{n}(\alpha_{i + 1}) - f(\alpha_{i + 1})| + |f(\alpha_{i + 1}) -
f(\bar{x})|\\ & < 2\varepsilon/5 \end{align*} where the first and third
terms in the sum is $< \varepsilon/10$ since $n \geq N$ and the second and
fourth terms in the sum is $< \varepsilon/10$ since $\bar{x}$ is between
$\alpha_{i}$ and $\alpha_{i + 1}$ and $|\alpha_{i + 1} - \alpha_{i}| <
\delta/2$ (and hence apply uniform continuity of $f$).
Therefore $f_{n} \rightarrow f$ uniformly.