What you want is use the $\varepsilon$-$\delta$ definition of continuity/semi-continuity
$$
\forall \varepsilon > 0, \quad \exists \delta > 0 \quad \text{s.t.} \quad |x-x_0| < \delta \quad \Longleftrightarrow \quad |f(x) - f(x_0)| < \varepsilon.
$$
Since $|f(x) - f(x_0)| < \varepsilon$ is equivalent to $f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon$, this statement is equivalent to
$$
\forall \varepsilon > 0, \quad \exists \delta > 0 \quad \text{s.t.} \quad |x-x_0| < \delta \quad \Longleftrightarrow \quad f(x_0) - \varepsilon < f(x) < f(x_0) + \varepsilon
$$
which means continuity is equivalent to upper and lower semi-continuity.
ADDED : I'll assume the definition of semi-continuity is the following, i.e. that the preimage of an open set of the form $\{ y \, | \, y > a \}$ is open for any upper semi-continuous function, and that the preimage of an open set of the form $\{ y \, | \, y < a \}$ is open for any lower semi-continuous function. If this is not the definition you have, let me know. I'm just guessing those definitions from the $\varepsilon$-$\delta$ definitions of continuity.
We show that continuity $\Longleftrightarrow$ upper and lower semi-continuity.
$(\Longrightarrow)$ This one is clear, since if the pre-image of any open set is open, then in particular are the pre-images of those of the form $(a, \infty)$ and $(-\infty,a)$.
$(\Longleftarrow)$ If $f^{-1}((a,\infty))$ and $f^{-1}((-\infty,b))$ are open sets, then since $f^{-1}((a,b)) = f^{-1}((a,\infty)) \cap f^{-1}((-\infty,b))$ and that the intersection of two open sets is open, then $f^{-1}((a,b))$ is open. Now any open set $\mathcal O$ in $\mathbb R$ is of the form
$$
\bigcup_{n=0}^{\infty} (a_n, b_n)
$$
where all the intervals $(a_n,b_n)$ are pairwise disjoint and $a_n \in \mathbb R \cup \{-\infty\}$, $b_n \in \mathbb R \cup \{\infty\}$. But
$$
f^{-1} \left( \bigcup_{i \in I} \mathcal O_i\right) = \bigcup_{i \in I} f^{-1} (\mathcal O_i)
$$
for any collection of sets indexed by any set $I$ (prove this trivially by the definition of pre-images and show $\subseteq$/$\supseteq$), thus,
$$
f^{-1}(\mathcal O) = \bigcup_{n=0}^{\infty} f^{-1}((a_n,b_n))
$$
which is an open set.
Hope that helps,
An example of a discontinuous finite-valued convex function on an (incomplete) normed linear space, posted by user127096.
Let $c_{00}$ be the subspace of $\ell^2$ consisting of the sequences with finitely many nonzero terms, with the norm inherited from $\ell^2$. The function $\varphi(x) = \sum_{n=1}^\infty n^2 x_n^2$ is convex on $c_{00}$ and takes values in $\mathbb R$ (no infinite values), yet is not continuous. Indeed, the sequence $n^{-1}e_n$ converges to $0$ in the norm of $X$, but $\varphi(n^{-1}e_n) = 1 \not\to 0 = \varphi(0)$. In particular, it is not lower semicontinuous. Also not locally bounded, which is related.
Best Answer
One definition that can be used for $\small\begin{array}{c}\text{upper}\\\text{lower}\end{array}$-semicontinuity is that $f$ is $\small\begin{array}{c}\text{upper}\\\text{lower}\end{array}$-semicontinuous if and only if $$ \{x:f(x)\lessgtr\alpha\} $$ is open for all $\alpha$.
Hints
Note that $$ \{x:\sup_{n\ge1}f_n(x)\gt\alpha\}=\bigcup_{n=1}^\infty \{x:f_n(x)\gt\alpha\} $$
One definition that can be used for continuity is that $f$ is continuous if and only if $f^{-1}(U)$ is open for all open $U$. Then note that $\{x:f(x)\gt\alpha\}=f^{-1}\left(\{x:x\gt\alpha\}\right)$.
In a fashion similar to 2. we can show that every continuous function is upper-semicontinuous. Thus, we just need to show that each function that is both upper and lower semicontinuous is continuous. Suppose that $f$ is both upper and lower semicontinuous. Then $$ f^{-1}(\alpha,\beta)=\{x:f(x)\gt\alpha\}\cap\{x:f(x)\lt\beta\} $$ is open for all $(\alpha,\beta)$. Furthermore, for every open set, $U$, $$ U=\bigcup_{u\in U}(u-\epsilon_u,u+\epsilon_u) $$ where $\epsilon_u\gt0$ is chosen so that $(u-\epsilon_u,u+\epsilon_u)\subset U$.